NEET Physics Class 11 Chapter 3 Centre Of Mass Notes

Centre Of Mass

Every physical system has associated with it a certain point whose motion characterizes the motion of the whole system.

• When the system moves under some external forces, then this point moves as if the entire mass of the system is concentrated at this point and also the external force is applied at this point for translational motion.
• This point is called the center of mass of the system.

Centre Of Mass Of A System Of ‘N’ Discrete Particles

Consider a system of N point masses m₁, m₂, m₃, ……………. mn whose position vectors from origin O are given by \(\vec{r}_1, \vec{r}_2, \vec{r}_3, \ldots \ldots \ldots \ldots. \vec{r}_n\) respectively. Then the position vector of the center of mass C of the system is given by.

⇒ \(\vec{r}_{c m}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2+\ldots \ldots . .+m_n \vec{r}_n}{m_1+m_2+\ldots \ldots . .+m_n}\)

⇒ \(\vec{r}_{c m}=\frac{\sum_{i=1}^n m_i \vec{r}_i}{\sum_{i=1}^n m_i}\)

⇒ \(\vec{r}_{\mathrm{cm}}=\frac{1}{M} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{m}_{\mathrm{i}} \vec{r}_{\mathrm{i}}\)

where, \(\mathrm{m}_{\mathrm{i}} \vec{r}_{\mathrm{i}}\) iis called the moment of mass of the particle w.r.t O.

⇒ \(M=\left(\sum_{i=1}^n m_i\right)\) is the total mass of the system.

Note: If the origin is taken at the center of mass then \(\sum_{i=1}^n m_i \vec{r}_i=0\). Hence, the COM is the point about which the sum of the “mass moments” of the system is zero.

Position Of Com Of Two Particles

The Centre of mass of two particles of masses m₁ and m₂ separated by a distance r lies in between the two particles. The distance of the center of mass from any of the particles (r) is inversely proportional to the mass of the particle (m)

i.e. \(r \propto 1 / m\)

⇒ \(\frac{r_1}{r_2}=\frac{m_2}{m_1}\)

⇒ \(m_1 r_1=m_2 r_2\)

⇒ \(r_1=\left(\frac{m_2}{m_2+m_1}\right) r \text { and } r_2=\left(\frac{m_1}{m_1+m_2}\right) r\)

Here, r₁= distance of COM from m₁

and r₂= distance of COM from m₂

From the above discussion, we see that

r₁= r₂= 1/2 if m1= m₂, i.e., COM lies midway between the two particles of equal masses.

Similarly, r₁> r₂ if  m₁< m₂ and r₁< r₂if m₂<₁m1, i.e., COM is nearer to the particle having a larger mass.

Question 1. Two particles of mass 1 kg and 2 kg are located at x = 0 and x = 3 m. Find the position of their center of mass.
Answer:

Since both the particles lie on the x-axis, the COM will also lie on the x-axis. Let the COM be located at x = x, then

r₁= distance of COM from the particle of mass 1 kg = x

and r₂ = distance of COM from the particle of mass 2 kg = (3 – x)

Using \(\frac{r_1}{r_2}=\frac{m_2}{m_1}\)

or \(\frac{x}{3-x}=\frac{2}{1} \text { or } x=2 m\)

Thus, the COM of the two particles is located at x = 2 m.

Question 2. The position vector of three particles of masses m1= 1 kg, m2= 2 kg and m3= 3 kg are

⇒ \(\vec{r}_1=(\hat{i}+4 \hat{j}+\hat{k}) m, \vec{r}_2=(\hat{i}+\hat{j}+\hat{k}) m\) and \(\vec{r}_3=(2 \hat{i}-\hat{j}-2 \hat{k}) m\) respectively.

Find the position vector of their center of mass.

Answer:

The position vector of COM of the three particles will be given by \(\vec{r}_{\text {COM }}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2+m_3 \vec{r}_3}{m_1+m_2+m_3}\)

Substituting the values, we get

⇒ \(\overrightarrow{\mathrm{r}}_{\text {COM }}=\frac{(1)(\hat{i}+4 \hat{j}+\hat{k})+(2)(\hat{i}+\hat{j}+\hat{k})+(3)(2 \hat{i}-\hat{j}-2 \hat{k})}{1+2+3}\)

⇒ \(\frac{1}{2}(3 \hat{i}+\hat{j}-\hat{k}) m\)

Question 3. Four particles of mass 1 kg, 2 kg, 3 kg, and 4 kg are placed at the four vertices A, B, C, and D of a square of side 1 m. Find the position of the center of mass of the particles.

Answer:

Assuming D as the origin, DC as the x-axis, and DA as the y-axis, we have

m₁= 1 kg, (x₁, y₁) = (0, 1m)

m₂= 2 kg, (x₂, y₂) = (1m, 1m)

m₃= 3 kg, (x₃, y₃) = (1m, 0)

and m₄= 4 kg, (x₄, y₄) = (0, 0)

Co-ordinates of their COM are

⇒ \(x_{\text {com }}=\frac{m_1 x_1+m_2 x_2+m_3 m_3+m_4 x_4}{m_1+m_2+m_3+m_4}\)

⇒ \(\frac{(1)(0)+2(1)+3(1)+4(0)}{1+2+3+4}\)

⇒ \(\frac{5}{10}\)

⇒ \(\frac{1}{2}\)

= 0.5m

Similarly, \(\mathrm{y}_{\text {com }}=\frac{\mathrm{m}_1 \mathrm{y}_1+\mathrm{m}_2 \mathrm{y}_2+\mathrm{m}_3 \mathrm{y}_3+\mathrm{m}_4 \mathrm{y}_4}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3+\mathrm{m}_4}\)

⇒ \(\frac{(1)(1)+2(1)+3(0)+4(0)}{1+2+3+4}\)

⇒ \(\frac{3}{10}\)

0.3

∴ (x_COM, y_COM) = (0.5 m, 0.3 m)

Thus, the position of COM of the four particles is as shown in the figure.

Question 4. Consider a two-particle system with the particles having masses m₁ and m₁. If the first particle is pushed towards the center of mass through a distance d, by what distance should the second particle be moved to keep the center of mass at the same position?
Answer:

Consider figure. Suppose the distance of m1 from the center of mass C is x₁ and that of m₂ from C is x₂. Suppose the mass m₂ is moved through a distance d′ towards C to keep the center of mass at C.

Then, m₁x₁= m₂x₂………(1)

and m₁(x₁– d) = m₂(x₂– d′). ……… (2)

Subtracting from

m₁d = m₂d′

or, d′ = \(\frac{m_1}{m_2} d\)

Centre Of Mass Of A Continuous Mass Distribution

For continuous mass distribution, the center of mass can be located by replacing the summation sign with an integral sign. Proper limits for the integral are chosen according to the situation

⇒ \(x_{c m}=\frac{\int x d m}{\int d m}, y_{c m}=\frac{\int y d m}{\int d m}, z_{c m}=\frac{\int z d m}{\int d m}\)

∫dm= M (mass of the body)

⇒ \(\overrightarrow{\mathrm{r}}_{\mathrm{cm}}=\frac{1}{\mathrm{M}} \int \overrightarrow{\mathrm{r}} \mathrm{dm}\)

Note: If an object has symmetric mass distribution about the axis then the y coordinate of COM is zero and vice versa

Centre Of Mass Of A Uniform Rod

Suppose a rod of mass M and length L is lying along the x-axis with its one end at x = 0 and the other at x = L. Mass per unit length of the rod = \(\frac{M}{L}\)

Hence, dm, (the mass of the element dx situated at x = x is) = \(\frac{M}{L}\) dx

The coordinates of the element dx are (x, 0, 0). Therefore, the x-coordinate of COM of the rod will be

⇒ \(x_{\text {com }}=\frac{\int_0^L x d m}{\int d m}=\frac{\int_0^L(x)\left(\frac{M}{L} d x\right)}{M}\)

⇒ \(\frac{1}{L} \int_0^L x d x=\frac{L}{2}\)

The y-coordinate of COM is

⇒ \(y_{\text {com }}=\frac{\int y d m}{\int d m}=0\)

Similarly, Z_COM = 0

i.e., the coordinates of COM of the rod are \(\left(\frac{\mathrm{L}}{2}, 0,0\right)\) i.e. it lies at the center of the rod.

Question 1. A rod of length L is placed along the x-axis between x = 0 and x = L. The linear density (mass/length) λ of the rod varies with the distance x from the origin as λ = Rx. Here, R is a positive constant. Find the position of the center of mass of this rod.
Answer:

The mass of element dx situated at x = x is

dm = λ dx = Rx dx

The COM of the element has coordinates (x, 0, 0).

Therefore, the x-coordinate of COM of the rod will be

⇒ \(x_{\text {com }}=\frac{\int_0^L x d m}{\int d m}\)

⇒ \(\frac{\int_0^L(x)(R x) d x}{\int_0^L(R x) d x}\)

⇒ \(\frac{R \int_0^L x^2 d x}{R \int_0^L x d x}\)

⇒ \(\frac{\left[\frac{x^3}{3}\right]_0^L}{\left[\frac{x^2}{2}\right]_0^L}\)

⇒ \(\frac{2 L}{3}\)

The y-coordinate of COM of the rod is \(\mathrm{y}_{\mathrm{com}}=\frac{\int \mathrm{ydm}}{\int \mathrm{dm}}\) = 0 (as y = 0)

Similarly, Z_COM = 0

Hence, the center of mass of the rod lies at \(\left[\frac{2 L}{3}, 0,0\right]\)

1. The center of mass of a uniform rectangular, square, or circular plate lies at its center. Axis of symmetry plane of symmetry.

2. For a laminar type (2-dimensional) body with uniform negligible thickness the formulae for finding the position of the center of mass are as follows:

⇒ \(\vec{r}_{\text {COM }}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2+\ldots}{m_1+m_2+\ldots}=\frac{\rho A_1 t \vec{r}_1+\rho A_2 t_r+\ldots}{\rho A_1 t+\rho A_2 t+\ldots}\)

(∵ m = ρAt)

⇒ \(\vec{r}_{\text {COM }}=\frac{A_1 \vec{r}_1+A_2 \vec{r}_2+\ldots}{A_1+A_2+\ldots}\)

Here, A stands for the area,

3. If some mass of area is removed from a rigid body, then the position of the center of mass of the remaining portion is obtained from the following formulae:

⇒ \(\vec{r}_{\text {COM }}=\frac{m_1 \vec{r}_1-m_2 \vec{r}_2}{m_1-m_2}\)

or \(\overrightarrow{\mathrm{r}}_{\text {COM }}=\frac{\mathrm{A}_1 \overrightarrow{\mathrm{r}}_1-\mathrm{A}_2 \overrightarrow{\mathrm{r}}_2}{\mathrm{~A}_1-\mathrm{A}_2}\)

⇒ \(x_{\text {COM }}=\frac{m_1 x_1-m_2 x_2}{m_1-m_2}\)

or \(\mathrm{x}_{\text {COM }}=\frac{\mathrm{A}_1 \mathrm{x}_1-\mathrm{A}_2 \mathrm{x}_2}{\mathrm{~A}_1-\mathrm{A}_2}\)

⇒ \(\mathrm{y}_{\text {COM }}=\frac{\mathrm{m}_1 \mathrm{y}_1-\mathrm{m}_2 \mathrm{y}_2}{\mathrm{~m}_1-\mathrm{m}_2}\)

or \(\mathrm{y}_{\text {COM }}=\frac{\mathrm{A}_1 \mathrm{y}_1-\mathrm{A}_2 \mathrm{y}_2}{\mathrm{~A}_1-\mathrm{A}_2}\)

and \(z_{\text {COM }}=\frac{m_1 z_1-m_2 z_2}{m_1-m_2}\)

or \(z_{\text {COM }}=\frac{A_1 z_1-A_2 z_2}{A_1-A_2}\)

Here, m₁, A₁, x₁, y₁ and z₁ are the values for the whole mass while m₂, A₂, \(\vec{r}_2, \vec{x}_2\) y₂ and z₂are the values for the mass which has been removed. Let us see two Questions in support of the above theory.

Question 2. Find the position of the center of mass of the uniform lamina shown in the figure.

Answer:

Here,

A₁ = area of complete circle = πa²

A₂= area of small circle = \(\pi\left(\frac{a}{2}\right)^2=\frac{\pi a^2}{4}\)

(x₁, y₁) = coordinates of centre of mass of large circle = (0, 0)

and (x₂, y₂) = coordinates of centre of mass of small circle = \(\left(\frac{\mathrm{a}}{2}, 0\right)\)

Using \(x_{\text {COM }}=\frac{\mathrm{A}_1 \mathrm{x}_1-\mathrm{A}_2 \mathrm{x}_2}{\mathrm{~A}_1-\mathrm{A}_2}\)

we get \(\mathrm{x}_{\text {com }}=\frac{-\frac{\pi \mathrm{a}^2}{4}\left(\frac{\mathrm{a}}{2}\right)}{\pi \mathrm{a}^2-\frac{\pi \mathrm{a}^2}{4}}=\frac{-\left(\frac{1}{8}\right)}{\left(\frac{3}{4}\right)} \mathrm{a}=-\frac{\mathrm{a}}{6}\)

and y_COM = 0 as y₁ and y₂ both are zero.

Therefore, the coordinates of COM of the lamina shown in the figure are \(\left(-\frac{a}{6}, 0\right)\)

Centre Of Mass Of Some Common Systems

A system of two point masses m₁ r₁= m₂ r₂ The center of mass lies closer to the heavier mass.

Rectangular plate (By symmetry)

⇒ \(\mathrm{x}_{\mathrm{c}}=\frac{\mathrm{b}}{2}\)

⇒ \(y_c=\frac{L}{2}\)

A triangular plate (By qualitative argument)

at the centroid : \(y_c=\frac{h}{3}\)

A semi-circular ring \(y_c=\frac{2 R}{\pi}\)

xc = 0

A semi-circular disc

⇒ \(y_c=\frac{4 R}{3 \pi}\)

xc = 0

A hemispherical shell

⇒ \(y_c=\frac{R}{2}\)

xc = 0

A solid hemisphere

⇒ \(y_c=\frac{3 R}{8}\)

xc= 0

A circular cone (solid)

⇒ \(y_c=\frac{h}{4}\)

A circular cone (hollow)

⇒ \(y_c=\frac{h}{3}\)

Question 1. A uniform thin rod is bent in the form of closed loop ABCDEFA as shown in the figure. The y-coordinate of the center of mass of the system is

1. \(\frac{2 r}{\pi}\)
2. \(-\frac{6 r}{3 \pi+2}\)
3. \(-\frac{2 r}{\pi}\)
4. Zero

Answer:

The center of mass of a semicircular ring is at a distance from its center.

(Let λ = mass/length)

∴ \(Y_{c m}=\frac{\lambda \pi r \times \frac{2 r}{\pi}-\lambda \times 2 \pi r \times \frac{4 r}{\pi}}{\lambda \pi r+\lambda r+\lambda r+\lambda \times 2 \pi r}=-\frac{6 r}{3 \pi+2}\)

Motion Of Centre Of Mass And Conservation Of Momentum:

Velocity Of Centre Of Mass Of System

⇒ \(\vec{v}_{c m}=\frac{m_1 \frac{d \vec{r}_1}{d t}+m_2 \frac{d\vec{r}_2}{d t}+m_3 \frac{d \vec{r}_3}{d t} \ldots \ldots \ldots \ldots . .+m_n \frac{d \vec{r}_n}{d t}}{M}\)

⇒ \(\frac{m_1 \vec{v}_1+m_2 \vec{v}_2+m_3 \vec{v}_3 \ldots \ldots \ldots .+m_n \vec{v}_n}{M}\)

Here, the numerator of the right-hand side term is the total momentum of the system i.e., the summation of momentum of the individual component (particle) of the system

Hence, the velocity of the center of mass of the system is the ratio of the momentum of the system to the mass of the system.

∴ \(\overrightarrow{\mathrm{P}}_{\text {System }}=\mathrm{M} \overrightarrow{\mathrm{v}}_{\mathrm{cm}}\)

Acceleration Of Centre Of Mass Of System

⇒ \(\vec{a}_{c m}=\frac{m_1 \frac{d \overrightarrow{v_1}}{d t}+m_2 \frac{d \overrightarrow{v_2}}{d t}+m_3 \frac{d \overrightarrow{v_3}}{d t} \ldots \ldots \ldots \ldots . .+m_n \frac{d \overrightarrow{v_n}}{d t}}{M}\)

⇒ \(\frac{m_1 \vec{a}_1+m_2 \vec{a}_2+m_3 \vec{a}_3 \ldots \ldots \ldots . .+m_n \vec{a}_n}{M}\)

⇒ \(\frac{\text { Net forceonsystem }}{\mathrm{M}}\)

⇒ \(\frac{\text { Net External Force }+ \text { Net internal Force }}{\mathrm{M}}\)

⇒ \(\frac{\text { Net External Force }}{\mathrm{M}}\)

(∵ action and reaction both of an internal force must be within the system. Vector summation will cancel all internal forces and hence net internal force on the system is zero)

∴ \(\overrightarrow{\mathrm{F}}_{\mathrm{ext}}=\mathrm{M} \overrightarrow{\mathrm{a}}_{\mathrm{cm}}\)

Where \(\overrightarrow{\mathrm{F}}_{\text {ext }}\) is the sum of the ‘external’ forces acting on the system. The internal forces which the

particles exert on one another play absolutely no role in the motion of the center of mass.

If no external force is acting on a system of particles, the acceleration of center of mass of the system will be zero. If a_c= 0, it implies that v_c must be a constant and if v_cm is a constant, it implies that the total momentum of the system must remain constant. It leads to the principle of conservation of momentum in the absence of external forces.

If \(\overrightarrow{\mathrm{F}}_{\text {ext }}\) = 0 ext = then \(\overrightarrow{\mathrm{v}}_{\mathrm{cm}}\) = constant

“If a resultant external force is zero on the system, then the net momentum of the system must remain constant”.

Motion Of COM In A Moving System Of Particles:

COM at rest :

If F_ext = 0 and V_cm= 0, then COM remains at rest. Individual components of the system may move and have non-zero momentum due to mutual forces (internal), but the net momentum of the system remains zero.

All the particles of the system are at rest.

Particles are moving such that their net momentum is zero.

A bomb at rest suddenly explodes into various smaller fragments, all moving in different directions then, since the explosive forces are internal and there is no external force on the system for explosion therefore, the COM of the bomb will remain at the original position and the fragment fly such that their net momentum remains zero.

• Two men standing on a frictionless platform, push each other, then also their net momentum remains zero because the push forces are internal for the two-men system.
• A boat floating in a lake also has a net momentum of zero if the people on it change their position, because the friction force required to move the people is internal to the boat system.
• Objects initially at rest, if moving under mutual forces (electrostatic or gravitation)also have net momentum zero.
• A light spring of spring constant k is kept compressed between two blocks of masses m₁ and m₂ on a smooth horizontal surface. When released, the blocks acquire velocities in opposite directions, such that the net momentum is zero.

(In a fan, all particles are moving but COM is at rest

COM Moving With Uniform Velocity:

If F_ext = 0, then Vcm remains constant therefore, the net momentum of the system also remains conserved. Individual components of the system may have variable velocity and momentum due to mutual forces (internal), but the net momentum of the system remains constant and COM continues to move with the initial velocity.

1. All the particles of the system are moving with the same velocity.
2. For example: A car moving with uniform speed on a straight road, has its COM moving with a constant velocity.
3. Internal explosions/breaking does not change the motion of COM and net momentum remains conserved.
4. A bomb moving in a straight line suddenly explodes into various smaller fragments, all moving in different directions then, since the explosive forces are internal and there is no external force on the system for explosion therefore, the COM of the bomb will continue the original motion and the fragment fly such that their net momentum remains conserved.
5. Man jumping from a cart or buggy also exerts internal forces therefore net momentum of the system and hence, the Motion of COM remains conserved.
6. Two moving blocks connected by a light spring on a smooth horizontal surface. If the acting forces are only due to spring then COM will remain in its motion and momentum will remain conserved.
7. Particles colliding in the absence of external impulsive forces also have their momentum conserved.

COM Moving With Acceleration:

If an external force is present then COM continues its original motion as if the external force is acting on it, irrespective of internal forces.

COM Moving With Acceleration Example:

Projectile motion: An axe thrown in the air at an angle θ with the horizontal will perform a complicated motion of rotation as well as a parabolic motion under the effect of gravitation

⇒ \(\mathrm{H}_{\mathrm{com}}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}}\)

⇒ \(R_{c o m}=\frac{u^2 \sin 2 \theta}{g}\)

⇒ \(T=\frac{2 u \sin \theta}{g}\)

Circular Motion: A rod hinged at an end, rotates, then its COM performs circular motion. The centripetal force (Fc) required in the circular motion is assumed to be acting on the COM.

F_c = ω²RCOM

Question 1. A projectile is fired at a speed of 100 m/s at an angle of 37º above the horizontal. At the highest point, the projectile breaks into two parts of mass ratio 1 : 3, the lighter piece coming to rest. Find the distance from the launching point to the point where the heavier piece lands.
Answer:

Internal force does not affect the motion of the center of mass, the center of mass hits the ground at the position where the original projectile would have landed. The range of the original projectile is,

⇒ \(\mathrm{x}_{\text {com }}=\frac{2 \mathrm{u}^2 \sin \theta \cos \theta}{\mathrm{g}}\)

⇒ \(\frac{2 \times 10^4 \times \frac{3}{5} \times \frac{4}{5}}{10} \mathrm{~m}\)

= 960 m

The center of mass will hit the ground at this position. As the smaller block comes to rest after breaking, it falls down vertically and hits the ground at half of the range, i.e., at x = 480 m. If the heavier block hits the ground at x2, then

⇒ \(x_{\text {COM }}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\)

960 = \(\frac{(m)(480)+(3 m)\left(x_2\right)}{(m+3 m)}\)

x₂= 1120

Momentum Conservation:

The total linear momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its center of mass.

⇒ \(\overrightarrow{\mathrm{P}}=\mathrm{M} \overrightarrow{\mathrm{v}}_{\mathrm{cm}}\)

⇒ \(\vec{F}_{\text {ext }}=\frac{\overrightarrow{d P}}{\mathrm{dt}}\)

If \(\vec{F}_{\text {ext }}=0 \frac{\mathrm{dP}}{\mathrm{dt}} \Rightarrow=0\)

⇒ \(\vec{p}\) = constant

When the vector sum of the external forces acting on a system is zero, the total linear momentum of the system remains constant.

⇒ \(\vec{P}_1+\vec{P}_2+\vec{P}_3+\ldots \ldots \ldots \ldots \ldots+=\vec{P}_n\) constant

Question 2. A shell is fired from a cannon with a speed of 100 m/s at an angle 60º with the horizontal (positive x-direction). At the highest point of its trajectory, the shell explodes into two equal fragments. One of the fragments moves along the negative x-direction with a speed of 50 m/s. What is the speed of the other fragment at the time of the explosion?
Answer:

As we know in the absence of external force the motion of the center of mass of a body remains unaffected. Thus, here the center of mass of the two fragments will continue to follow the original projectile path. The velocity of the shell at the highest point of trajectory is

v_M= ucosθ = 100 ×cos60º = 50 m/s.

Let v₁ be the speed of the fragment that moves along the negative x-direction and the other fragment has speed v₂, which must be along the positive x-direction. Now from momentum conservation, we have

⇒ \(m v=\frac{-m}{2} v_1+\frac{m}{2} v_2\)

or 2v = v₂-v₁

or v₂ = 2v+v₁

= (2×50) + 50 = 150m/s

Question 3. A man of mass m is standing on a platform of mass M kept on smooth ice. If the man starts moving on the platform with a speed v relative to the platform, with what velocity relative to the ice does the platform recoil?
Answer:

Consider the situation shown in the figure. Suppose the man moves at a speed w towards the right and the platform recoils at a speed of V towards the left, both relative to the ice. Hence, the speed of the man relative to the platform is V + w. By the question,

V + w = v, or w = v – V ………….(1)

Taking the platform and the man to be the system, there is no external horizontal force on the system. The linear momentum of the system remains constant. Initially, both the man and the platform were at rest. Thus,

0 = MV – mw or, MV = m (v – V) [Using (1)]

or, V = \(\frac{\mathrm{mv}}{\mathrm{M}+\mathrm{m}}\)

Question 4. In a free space, a rifle of mass M shoots a bullet of mass m at a stationary block of mass M distance D away from it. When the bullet has moved through a distance d towards the block the center of mass of the bullet-block system is at a distance of :

1. \(\frac{(D-d) m}{M+m}\) from the block
2. \(\frac{m d+M D}{M+m}\) from the rifle
3. \(\frac{2 \mathrm{dm}+\mathrm{DM}}{\mathrm{M}+\mathrm{m}}\) from the rifle
4. \((D-d) \frac{M}{M+m}\) from the bullet

Answer: (1,2,4)

As; Mx = m(D – d – x) x = \(\frac{m(D-d)}{M+m}\) from the block

and x’ = D – d – x = \(\frac{(D-d) M}{M+m}\) from the bullet.

Question 5. The center of mass of two masses m & m′ moves by distance \(\frac{x}{5}\) when mass m is moved by distance x and m′ is kept fixed. The ratio \(\frac{m’}{m}\) is

1. 2
2. 4
3. 1/4
4. None of these

Answer: 2. 4

(m + m′)\(\frac{x}{5}\) = mx + m′O

∴ m + m′ = 5 m ; m’ = 4m

⇒ \(\frac{m’}{m}\) = 4

Question 6. A person P of mass 50 kg stands in the middle of a boat of mass 100 kg moving at a constant velocity of 10 m/s with no friction between water and boat and also the engine of the boat is shut off. With what velocity (relative to the boat’s surface) should the person move so that the boat comes to rest? Neglect friction between water and boat.

1. 30 m/s towards right
2. 20 m/s towards right
3. 30 m/s towards left
4. 20 m/s towards left

Answer: 1. 30 m/s towards right

The momentum of the system remains conserved as no external force is acting on the system in a horizontal direction.

∴ (50 + 100) 10 = 50 × V + 100 × 0 ⇒ V = 30 m/s towards right, as boat is at rest.

V = 30 m/s Pboat

Question 7. Two men of masses 80 kg and 60 kg are standing on a wood plank of mass 100 kg, that has been placed over a smooth surface. If both the men start moving toward each other with speeds 1 m/s and 2 m/s respectively then find the velocity of the plank by which it starts moving.

Answer:

Applying momentum conservation ;

(80) 1 + 60 (– 2) = (80 + 60 + 100) v

v = \(\frac{-40}{240}\)

⇒ \(-\frac{1}{6}\) m/sec.

Question 8. Each of the blocks shown in the figure has a mass of 1 kg. The rear block moves with a speed of 2 m/s towards the front block kept at rest. The spring attached to the front block is light and has a spring constant of 50 N/m. Find the maximum compression of the spring. Assume, on a frictionless surface

Answer:

Maximum compression will take place when the blocks move with equal velocity. As no net external horizontal force acts on the system of the two blocks, the total linear momentum will remain constant. If V is the common speed at maximum compression, we have,

(1 kg) (2 m/s) = (1 kg)V + (1 kg)V or, V = 1 m/s.

Initial kinetic energy = \(\frac{1}{2}\)(1 kg) (2 m/s)2 = 2 J.

Final kinetic energy = \(\frac{1}{2}\)(1 kg) (1m/s)2 + \(\frac{1}{2}\) (1 kg) (1 m/s)2 = 1 J

The kinetic energy lost is stored as the elastic energy in the spring.

Hence, \(\frac{1}{2}\)(50 N/m) x2 = 2J – 1J = 1 J

or x = 0.2 m.

Question 9. The figure shows two blocks of masses 5 kg and 2 kg placed on a frictionless surface and connected with a spring. An external kick gives a velocity of 14 m/s to the heavier block towards the lighter one. Find the velocity gained by the center of mass

Answer:

The velocity of the center of mass is

⇒ \(v_{c m}=\frac{5 \times 14+2 \times 0}{5+2}\)

= 10m/s

Question 10. The two blocks A and B of the same mass are connected to a spring and placed on a smooth surface. They are given velocities (as shown in the figure) when the spring is at its natural length:

1. The maximum velocity of B will be 10 m/s
2. The maximum velocity of B will be greater than 10 m/s
3. The spring will have maximum extension when A and B both stop
4. The spring will have maximum extension when both move toward the left.

Answer:

• Suppose B moves with a velocity more than 10 m/s a should move at a velocity greater than 5 m/s and increase the overall energy which is not possible since there is no external force acting on the system.
• Hence B should move with a maximum velocity of 10 m/s. Also, both A and B can never stop so as to keep the momentum constant.
• Also, both A and B can never move towards the left simultaneously for momentum to remain conserved. Hence only (A) is correct.

Impulse

Impulse of a force \(\overrightarrow{\mathrm{F}}\) acting on a body for the time interval t = t₁ to t = t₂ is defined as :- dv

⇒ \(\vec{I}=\int_{t_1}^{t_2} \vec{F} d t\)

⇒ \(\overrightarrow{\mathrm{I}}=\int \overrightarrow{\mathrm{F}} \mathrm{dt}=\int \mathrm{m} \frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}} \mathrm{dt}=\int \mathrm{md} \overrightarrow{\mathrm{v}}\)

⇒ \(\overrightarrow{\mathrm{I}}=\mathrm{m}\left(\overrightarrow{\mathrm{v}}_2-\overrightarrow{\mathrm{v}}_1\right)\)

ΔP= change in momentum due to force F

Also, \(\overrightarrow{\mathrm{I}}_{\text {Res }}=\int_{\mathrm{t}_1}^{\mathrm{t}_2} \overrightarrow{\mathrm{F}}_{\text {Res }} d t=\Delta \overrightarrow{\mathrm{P}}\)

(impulse-momentum theorem)

Note: Impulse applied to an object in a given time interval can also be calculated from the area under force time (F-t) graph in the same time interval.

Instantaneous Impulse:

There are many cases when a force acts for such a short time that the effect is instantaneous, for example., a bat striking a ball.

In such cases, although the magnitude of the force and the time for which it acts may each be unknown the value of their product (i.e., impulse) can be known by measuring the initial and final moment. Thus, we can write.

⇒ \(\overrightarrow{\mathrm{I}}=\int \overrightarrow{\mathrm{F}} \mathrm{dt}=\Delta \overrightarrow{\mathrm{P}}=\overrightarrow{\mathrm{P}}_{\mathrm{f}}-\overrightarrow{\mathrm{P}}_{\mathrm{i}}\)

Important Points:

1. It is a vector quantity.
2. Dimensions = [MLT^-1]
3. SΙ unit = kg m/s
4. Direction is along a change in momentum.
5. Magnitude is equal to the area under the F-t. graph.
6. \(\overrightarrow{\mathrm{I}}=\int \overrightarrow{\mathrm{F}}_{\mathrm{dt}}=\overrightarrow{\mathrm{F}}_{\mathrm{av}} \int \mathrm{dt}=\overrightarrow{\mathrm{F}}_{\mathrm{av}} \Delta \mathrm{t}\)
7. It is not a property of a particle, but it is a measure of the degree to which an external force changes the momentum of the particle.

Question 1. The hero of a stunt film fires 50 g bullets from a machine gun, each at a speed of 1.0 km/s. If he fires 20 bullets in 4 seconds, what average force does he exert against the machine gun during this period?
Answer:

The momentum of each bullet

= (0.050 kg) (1000 m/s) = 50 kg-m/s.

The gun has imparted this much amount of momentum with each bullet fired. Thus, the rate of change of momentum of the gun

⇒ \(\frac{(50 \mathrm{~kg}-\mathrm{m} / \mathrm{s}) \times 20}{4 \mathrm{~s}}\)

= 250 N

In order to hold the gun, the hero must exert a force of 250 N against the gun.

Collision Or Impact

A collision is an event in which an impulsive force acts between two or more bodies for a short time, which results in a change in their velocities.

Note:

1. In a collision, particles may or may not come in physical contact.
2. The duration of collision, Δt is negligible as compared to the usual time intervals of observation of motion.

The collision is in fact a redistribution of the total momentum of the particles. Thus, the law of conservation of linear momentum is indispensable in dealing with the phenomenon of collision between particles.

Line Of Impact

The line passing through the common normal to the surfaces in contact during impact is called a line of impact. The force during collision acts along this line on both bodies.

The direction of the Line of impact can be determined by:

1. The geometry of colliding objects like spheres, discs, wedges,s, etc.
2. Direction of change of momentum.

If one particle is stationary before the collision then the line of impact will be along its motion after collision.

Classification Of Collisions:

On The Basis Of The Line Of Impact

1. Head-on collision: If the velocities of the colliding particles are along the same line before and after the collision.
2. Oblique collision: If the velocities of the colliding particles are along different lines before and after the collision.

On The Basis Of Energy:

1. Elastic collision: In an elastic collision, the colliding particles regain their shape and size completely after the collision. i.e., no fraction of mechanical energy remains stored as deformation potential energy in the bodies. Thus, the kinetic energy of the system after collision is equal to the kinetic energy of the system before collision. Thus in addition to the linear momentum, kinetic energy also remains conserved before and after collision.
2. Inelastic collision: In an inelastic collision, the colliding particles do not regain their shape and size completely after collision. Some fraction of mechanical energy is retained by the colliding particles in the form of deformation potential energy. Thus, the kinetic energy of the particles after collision is not equal to that of before collision. However, in the absence of external forces, law of conservation of linear momentum still holds good.
3. Perfectly inelastic: If the velocity of separation along the line of impact just after collision becomes zero then the collision is perfectly inelastic. Collision is said to be perfectly inelastic if both the particles stick together after collision and move with the same velocity,

Note: Actually collisions between all real objects are neither perfectly elastic nor perfectly inelastic, it’s inelastic in nature.

Examples Of Line Of Impact And Collisions Based On Line Of Impact

Two balls A and B are approaching each other such that their centres are moving along line CD.

Head on Collision

Two balls A and B are approaching each other such that their center is moving along dotted lines as shown in the figure.

Oblique Collision

Ball is falling on a stationary wedge.

Oblique Collision

Coefficient Of Restitution (e)

e = \(\frac{\text { Velocity of separation along line of impact }}{\text { Velocity of approach along line of impact }}\)

The most general expression for the coefficient of restitution is

e = \(\frac{\text { velocity of separation of points of contact along line of impact }}{\text { velocity of approach of point of contact along line of impact }}\)

Example For Calculation Of e:

Two smooth balls A and B approach each other such that their centers are moving along line CD in the absence of external impulsive force. The velocities of A and B just before collision are u₁ and u₂ respectively. The velocities of A and B just after collision are v₁ and v₂ respectively.

∵ F_ext = 0 momentum is conserved for the system.

⇒ m₁u₁+ m₂u₂= (m₁+ m₂)v = m₁v₁+ m₂v₂

⇒ v = \(\frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{m_1 v_1+m_2 v_2}{m_1+m_2}\)…..(1)

Impulse of Deformation:

JD = change in momentum of any one body during deformation.

= m₂(v – u₂) for m₂

= m₁(–v + u₁) for m₁

Impulse Of Reformation:

JR = change in momentum of any one body during Reformation.

= m₂(v₂– v) for m₂

= m₁(v – v₁) for m₁

e = \(=\frac{\text { Impulse of Reformation }\left(\overrightarrow{\mathrm{J}}_{\mathrm{R}}\right)}{\text { Impulse of Deformation }\left(\overrightarrow{\mathrm{J}}_{\mathrm{D}}\right)}\)

⇒ \(\frac{v_2-v_1}{u_1-u_2}\)

⇒ \(\frac{\text { Velocity of separation along line of impact }}{\text { Velocity of approach along line of impact }}\)

Note: e is independent of the shape and mass of the object but depends on the material. The coefficient of restitution is constant for a pair of materials.

1. e = 1 Velocity of separation along the LOI = Velocity of approach along the LOI Kinetic energy of particles after collision may be equal to that of before collision. Collision is elastic.
2. e = 0 Velocity of separation along the LOI = 0 Kinetic energy of particles after collision is not equal to that of before collision. Collision Is Perfectly Inelastic.
3. 0 < e < 1 Velocity of separation along the LOI < Velocity of approach along the LOI Kinetic energy of particles after collision is not equal to that of before collision. Collision is Inelastic.

Note: In case of contact collisions e is always less than unity.

∴ 0 ≤ e ≤ 1

Collision In One Dimension (Head on)

⇒ \(u_1>u_2\)

⇒ \(v_2>v_1\)

⇒ \(e=\frac{v_2-v_1}{u_1-u_2}\)

⇒ \(\left(u_1-u_2\right) e=\left(v_2-v_1\right)\)

By momentum conservation,

m₁u₁+ m₂u₂= m₁v₁+ m₂v₂

v₂= v₁+ e(u₁– u₂)

and \(v_1=\frac{m_1 u_1+m_2 u_2-m_2 e\left(u_1-u_2\right)}{m_1+m_2}\)

⇒ \(v_2=\frac{m_1 u_1+m_2 u_2+m_1 e\left(u_1-u_2\right)}{m_1+m_2}\)

Special Case:

e = 0

v₁= v₂

For perfectly inelastic collision, both the bodies, move with the same vel. after collision.

e = 1

and m₁= m₂= m,

we get v₁= u₂ and v₂= u₁

i.e., when two particles of equal mass collide elastically and the collision is head-on, they exchange their velocities., for example.,

m₁>> m₂

⇒ \(\mathrm{m}_1+\mathrm{m}_2 \approx \mathrm{m}_1 \text { and } \frac{\mathrm{m}_2}{\mathrm{~m}_1} \approx 0\)

v₁ = u₁ No change

and v₂= u₁+ e(u₁– u₂)

Now If e = 1

v₂= 2u₁– u₂

Question 1. Two identical balls are approaching each other on a straight line with velocities of 2 m/s and 4 m/s respectively. Find the final velocities, after elastic collision between them.

Answer:

The two velocities will be exchanged and the final motion is the reverse of the initial motion for both.

Question 2. Three balls A, B, and C of the same mass ‘m’ are placed on a frictionless horizontal plane in a straight line as shown. Ball A is moved with velocity u towards the middle ball B. If all the collisions are elastic then, find the final velocities of all the balls.

Answer:

A collides elastically with B and comes to rest but B starts moving with velocity u

After a while B collides elastically with C and comes to rest but C starts moving with velocity u

∴ Final velocities VA = 0; VB= 0 and VC= u

Question 3. Four identical balls A, B, C, and D are placed in a line on a frictionless horizontal surface. A and D are moved with the same speed ‘u’ towards the middle as shown. Assuming elastic collisions, find the final velocities.

Answer:

A and D collide elastically with B and C respectively and come to rest but B and C start moving with velocity u towards each other as shown

B and C collide elastically and exchange their velocities to move in opposite directions

Now, B and C collide elastically with A and D respectively and come to rest but A and D start moving with velocity u away from each other as shown

∴ Final velocities V_A = u (←); V₂= 0; V_C= 0 and V_D = u (→)

Question 4. Two particles of mass m and 2m moving in opposite directions on a frictionless surface collide elastically with velocity v and 2v respectively. Find their velocities after a collision, also find the fraction of kinetic energy lost by the colliding particles.

Answer:

Let the final velocities of m and 2m be v₁ and v₂ respectively as shown in the figure:

By conservation of momentum:

m(2v) + 2m(–v) = m(v₁) + 2m (v₂)

or 0 = mv₁+ 2mv₂ or v₁+ 2v₂= 0 ………(1)

and since the collision is elastic:

v₂ – v₁ = 2v –(–v) or v₂ – v₁ = 3v ………(2)

Solving the above two equations, we get,

v₂= v and v₁= –2v

i.e., the mass 2m returns with velocity v while the mass m returns with velocity 2v in the direction shown in figure:

The collision was elastic therefore, no kinetic energy is lost, KE loss = KE_i– KE_f

or \(\left(\frac{1}{2} m(2 v)^2+\frac{1}{2}(2 m)(-v)^2\right)-\left(\frac{1}{2} m(-2 v)^2+\frac{1}{2}(2 m) v^2\right)\) = 0

Question 5. On a frictionless surface, a ball of mass m moving at a speed v makes a head-on collision with an identical ball at rest. The kinetic energy of the balls after the collision is 3/4th of the original. Find the coefficient of restitution.
Answer:

As we have seen in the above discussion, that under the given conditions :

By using conservation of linear momentum and equation of e, we get,

⇒ \(v_1^{\prime}=\left(\frac{1+e}{2}\right) v\)

and \(v_2^{\prime}=\left(\frac{1-e}{2}\right) v\)

Given that \(\mathrm{K}_{\mathrm{f}}=\frac{3}{4} \mathrm{~K}_{\mathrm{i}}\)

or \(\frac{1}{2} m v_1^{\prime 2}+\frac{1}{2} m v_2^{\prime 2}=\frac{3}{4}\left(\frac{1}{2} m v^2\right)\)

Substituting the value, we get

⇒ \(\left(\frac{1+e}{2}\right)^2+\left(\frac{1-e}{2}\right)^2\)

⇒ \(\frac{3}{4}\)

or \(e=\frac{1}{\sqrt{2}}\)

Question 6. A block of mass 2 kg is pushed toward a very heavy object moving with 2 m/s closer to the block (as shown). Assuming elastic collision and frictionless surfaces, find the final velocities of the blocks.

Answer:

Let v₁ and v₂ be the final velocities of the 2kg block and heavy object respectively then,

v₁= u₁+ 1 (u₁– u₂) = 2u₁– u₂ = –14 m/s

v₂= –2m/s

Question 7. A ball is moving with velocity of 2 m/s towards a heavy wall moving towards the ball with a speed 1m/s as shown in fig. Assuming collision to be elastic, find the velocity of the ball immediately after the collision.

Answer:

The speed of wall will not change after the collision. So, let v be the velocity of the ball after collision in the direction shown in figure. Since collision is elastic (e = 1),

separation speed = approach speed

or v – 1 = 2 + 1 or v = 4 m/s

Collision In Two Dimensions (oblique)

Question 8. A ball of mass m hits a floor with a speed v0 making an angle of incidence a with the normal. The coefficient of restitution is e. Find the speed of the reflected ball and the angle of reflection of the ball.
Answer:

The component of velocity v₀ along common tangential direction v₀ sin α will remain unchanged. Let v be the component along a common normal direction after collision. Applying, Relative speed of separation = e (Relative speed of approach) along a common normal direction, we get v = ev₀cos α

Thus, after collision components of velocity v’ are v₀ sin α and ev₀ cos α

⇒ \(v^{\prime}=\sqrt{\left(v_0 \sin \alpha\right)^2+\left(e v_0 \cos \alpha\right)^2}\)

and tan β = \(\frac{v_0 \sin \alpha}{e v_0 \cos \alpha}\)

α or tan β = \(\frac{\tan \alpha}{e}\)

Note: For elastic collision, e = 1

∴ v’ = v₀ and β = α

Question 9. A ball of mass m makes an elastic collision with another identical ball at rest. Show that if the collision is oblique, the bodies go at right angles to each other after a collision.
Answer:

In a head-on elastic collision between two particles, they exchange their velocities. In this case, the component of ball 1 along the common normal direction, v cos θ becomes zero after a collision, while that of 2 becomes v cos θ.

While the components along the common tangent direction of both the particles remain unchanged. Thus, the components along the common tangent and common normal direction of both the balls in tabular form are given below.

From the above table and figure, we see that both the balls move at the right angle after collision with velocities v sin θ and v cos θ.

Note: When two identical bodies have an oblique elastic collision, with one body at rest before the collision, then the two bodies will go in ⊥different directions.

Variable Mass System

If a mass is added or ejected from a system, at rate μ kg/s and relative velocity \(\overrightarrow{\mathrm{v}}_{\mathrm{rel}}\)(w.r.t. the system), then the force exerted by this mass on the system has magnitude \(\)

Thrust Force \(\left(\vec{F}_t\right)\)

⇒\(\overrightarrow{\mathrm{F}}_{\mathrm{t}}=\overrightarrow{\mathrm{v}}_{\mathrm{rel}}\left(\frac{\mathrm{dm}}{\mathrm{dt}}\right)\)

Suppose at some moment t = t mass of a body is m and its velocity is \(\vec{v}_r\). After some time at t = t + dt its mass becomes (m – dm) and velocity becomes.

The mass dm is ejected with relative velocity \(\mu\left|\vec{v}_{\text {rel }}\right|\)

Absolute velocity of mass ‘dm’ is therefore \(\left(\vec{v}+\vec{v}_r\right)\). If no external forces are acting on the system, the r linear momentum of the system will remain conserved, or

⇒ \(\vec{P}_{\mathrm{i}}=\vec{P}_{\mathrm{f}}\)

or \(m \vec{v}=(m-d m)(\vec{v}+d \vec{v})+d m\left(\vec{v}+\vec{v}_r\right)\)

or \(m \vec{v}=m \vec{v}+m d \vec{v}-(d m) \vec{v}-(d m)(d \vec{v})+(d m) \vec{v}+\vec{v} d m\)

The term (dm)\((\mathrm{d} \overrightarrow{\mathrm{v}} \text { ) }\) is too small and can be neglected.

∴ \(\mathrm{m}\left(\frac{\mathrm{d} \overrightarrow{\mathrm{v}}}{\mathrm{dt}}\right)=\overrightarrow{\mathrm{v}}_{\mathrm{r}}\left(-\frac{\mathrm{dm}}{\mathrm{dt}}\right)\)

Here, \(\mathrm{m}\left(-\frac{\mathrm{d} \vec{v}}{d \mathrm{dt}}\right)=\text { thrust force }\left(\vec{F}_{\mathrm{t}}\right)\)

and \(-\frac{\mathrm{dm}}{\mathrm{dt}}\) = rate at which mass is ejecting or \(\vec{F}_t=\vec{v}_r\left(\frac{d m}{d t}\right)\)

Rocket Propulsion:

Let m₀ be the mass of the rocket at time t = 0. m its mass at any time t and v its velocity at that moment. Initially, let us suppose that the velocity of the rocket is u.

Further, let \(\left(\frac{-\mathrm{dm}}{\mathrm{dt}}\right)\) be the mass of the gas ejected per unit time and vrthe exhaust velocity of the gases with respect to rocket.

Usually \(\left(\frac{-\mathrm{dm}}{\mathrm{dt}}\right)\) and vr are kept constant throughout the journey of the dt rocket. Now, let us write few equations which can be used in the problems of rocket propulsion. At time t = t,

1. Thrust force on the rocket Ft= vrdm \(F_t=v_r\left(\frac{-d m}{d t}\right)\)(upwards)

2. Weight of the rocket W = mg (downwards)

3. Net force on the rocket Fnet = Ft– W (upwards)

or \(F_{\text {net }}=v_r\left(\frac{-d m}{d t}\right)-m g\)

4. Net acceleration of the rocket a = \(\frac{F}{m}\)

⇒ \(\frac{d v}{d t}=\frac{v_r}{m}\left(\frac{-d m}{d t}\right)-g\)

or \(\mathrm{dv}=\frac{v_r}{m}(-d m)-g d t\)

or \(\int_u^v d v=v_r \int_{m_0}^m \frac{-d m}{m}-g \int_0^t d t\)

Thus, v = u – gt + vr ln \(\left(\frac{m_0}{m}\right)\)…(i) m

Note:

1. Ft= vr \(\left(-\frac{\mathrm{dm}}{\mathrm{dt}}\right)\) is upwards, as v_r is downwards and \(\frac{\mathrm{dm}}{\mathrm{dt}}\) is negative.
2. If gravity is ignored and initial velocity of the rocket u = 0, Eq. (1) reduces to v = v_r ln \(\left(\frac{m_0}{m}\right)\)

Question 1. A rocket, with an initial mass of 1000 kg, is launched vertically upwards from rest under gravity. The rocket burns fuel at the rate of 10 kg per second. The burnt matter is ejected vertically downwards with a speed of 2000 ms^-2 relative to the rocket. If the burning stops after one minute. Find the maximum velocity of the rocket. (Take g as at 10 ms^-2)
Answer:

Using the velocity equation

v = u – gt + v_r ln\(\left(\frac{m_0}{m}\right)\)

Here u = 0, t = 60s, g = 10 m/s², vr= 2000 m/s, m₀= 1000 kg

and m = 1000 – 10 × 60 = 400 kg

We get v = 0 – 600 + 2000 ln \(\left(\frac{1000}{400}\right)\)

or v = 2000 ln 2.5 – 600

The maximum velocity of the rocket is 200(10 ln 2.5 – 3) = 1232.6 ms^-1

Linear Momentum Conservation In The Presence Of External Force.

⇒ \(\overrightarrow{\mathrm{F}}_{\mathrm{ext}}=\frac{\mathrm{d} \overrightarrow{\mathrm{P}}}{\mathrm{dt}}\)

⇒ \(\vec{F}_{\text {ext }} d t=d \vec{P}\)

⇒ \(\left.\mathrm{d} \overrightarrow{\mathrm{P}}=\overrightarrow{\mathrm{F}}_{\text {ext }}\right)_{\text {mpulsive }} \mathrm{dt}\)

∴ If \(\left.\vec{F}_{\text {ext }}\right)_{\text {mpuulise }}=0\)

⇒ \(\mathrm{d} \overrightarrow{\mathrm{P}}=0\)

or \(\vec{P}\) is constant 94

Note: Momentum is conserved if the external force present is non-impulsive. eg. gravitation or spring force

Question 1. Two balls are moving toward each other on a vertical line that collides with each other as shown. Find their velocities just after the collision.

Answer:

Let the final velocity of 4 kg ball just after collision be v. Since, an external force is gravitational which is non – impulsive, hence, linear momentum will be conserved.

Applying linear momentum conservation:

2(–3) + 4= 2+ 4(v) or v = \(\frac {1}{2}\)m/s

Question 2. Three particles of masses 0.5 kg, 1.0 kg, and 1.5 kg are placed at the three corners of a right-angled triangle of sides 3.0 cm, 4.0 cm, and 5.0 cm as shown in the figure. Locate the center of mass of the system.

Answer:

taking the x and y axes as shown.

Coordinates of body A = (0,0)

Coordinates of body B = (4,0)

Coordinates of body C = (0,3)

x – coordinate of c.m. = \(\frac{m_A x_A+m_B x_B+M_C r_C}{m_A+m_B+m_C}\)

⇒ \(\frac{0.5 \times 0+1.0 \times 4+1.5 \times 0}{0.5+1.0+1.5}\)

⇒ \(\frac{4}{3} \frac{\mathrm{cm}}{\mathrm{kg}}\)

= 1.33cm

similarly y – coordinates of c.m. = \(\frac{0.5 \times 0+1.0 \times 0+1.5 \times 3}{0.5+1.0+1.5}\)

⇒ \(\frac{4.5}{3}\)

= 1.5 cm

So, a center of mass is 1.33 cm right and 1.5 cm above particle A.

Question 3. A block A (mass = 4M) is placed on the top of a wedge B of base length l (mass = 20 M) as shown in the figure. When the system is released from rest. Find the distance moved by wedge B till block A reaches the lowest end of the wedge. Assume all surfaces are frictionless.

Answer:

The initial position of the center of mass

⇒ \(\frac{X_B M_B+X_A M_A}{M_B+M_B}\)

⇒ \(\frac{X_B \cdot 20 M+\ell .4 M}{24 M}\)

⇒ \(\frac{5 \mathrm{X}_{\mathrm{B}}+\ell}{6}\)

The final position of the center of mass

⇒ \(\frac{\left(X_B+x\right) 20 M+4 M x}{24 M}\)

⇒ \(\frac{5\left(X_B+x\right)+x}{6}\)

since there is no horizontal force on the system center of mass initially = center of mass finally.

5X_B+ l = 5X_B+ 5x + x

l = 6x

⇒ \(\frac{\ell}{6}\)

Question 4. An isolated particle of mass m is moving in a horizontal xy plane, along the x-axis. At a certain height above ground, it suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragment is at y = + 15 cm. Find the position of the heavier fragment at this instant.
Answer:

As the particle is moving along the x-axis, so, the y-coordinate of COM is zero.

⇒ \(Y_M M=Y_{\frac{M}{4}}\left(\frac{M}{4}\right)+Y_{\frac{M M}{4}}\left(\frac{3 M}{4}\right)\)

⇒ \(0 \times M=15\left(\frac{M}{4}\right)+Y_{\frac{3 M}{4}}\left(\frac{3 M}{4}\right)\)

⇒ \(\frac{Y_{3 M}}{4}=-5 \mathrm{~cm}\)

Question 5. A shell at rest at origin explodes into three fragments of masses 1 kg, 2 kg, and m kg. The fragments of masses 1 kg and 2 kg fly off with speeds of 12 m/s along the x-axis and 8 m/s along the axis respectively. If m kg flies off with speed 40 m/s then find the total mass of the shell.
Answer:

As initial velocity = 0, Initial momentum = (1 + 2 + m) × 0 = 0

Finally, let velocity of M = \(\vec{v}\)

We know \(|\vec{V}|\) = 40 m/s.

Initial momentum = final momentum

0 = \(1 \times 12 \hat{i}+2 \times 8 \hat{j}+m \vec{V}\)

⇒ \(\vec{V}=\frac{(12 \hat{i}+16 \hat{j})}{m}\)

⇒ \(|\vec{V}|=\sqrt{\frac{(12)^2+(16)^2}{m^2}}\)

⇒ \(\frac{1}{m} \sqrt{(12)^2+(16)^2}\) = 40 {given}

⇒ \(m=\sqrt{\frac{(12)^2+(16)^2}{40}}\)

= 0.5 kg Total mass = 1 + 2 + 0.5 = 3.5 kg

Question 6. A block moving horizontally on a smooth surface with a speed of 20 m/s bursts into two equal parts continuing in the same direction. If one of the parts moves at 30 m/s, with what speed does the second part move and what is the fractional change in the kinetic energy of the system?
Answer:

Applying momentum conservation:

⇒ \(m \times 20=\frac{m}{2} V+\frac{m}{2} \times 30\)

⇒ \(20=\frac{V}{2}+15\)

So, V = 10 m/s

initial kinetic energy = \(\frac{1}{2}\)m × (20)2 = 200 m

final kinetic energy = \(\frac{1}{2} \cdot \frac{m}{2} \cdot(10)^2+\frac{1}{2} \times \frac{m}{2}(30)^2\)

= 25 m + 225 m = 250 m

fractional change in kinetic energy = \(\frac{(\text { final K. E) }- \text { (initial K. E) }}{\text { initial K.E }}\)

⇒ \(\frac{250 m-200 m}{200 m}\)

⇒ \(\frac{1}{4}\)

Question 7. A block at rest explodes into three equal parts. Two parts start moving along the X and Y axes respectively with equal speeds of 10 m/s. Find the initial velocity of the third part.
Answer:

Let total mass = 3 m,

initial linear momentum = 3m × 0

Let velocity of third part = \(\overrightarrow{\mathrm{V}}\)

Using conservation of linear momentum:

⇒ \(m \times 10 \hat{i}+m \times 10 \hat{j}+m \vec{V}=0\)

So, \(\vec{V}=(-10 \hat{i}-10 \hat{j}) \mathrm{m} / \mathrm{sec} .\)

⇒ \(|\vec{V}|=\sqrt{(10)^2+(10)^2}=10 \sqrt{2}\) making angle 135o below x-axis

Question 8. Blocks A and B have masses of 40 kg and 60 kg respectively. They are placed on a smooth surface and the spring connected between them is stretched by 1.5m. If they are released from rest, determine the speeds of both blocks at the instant the spring becomes unstretched.

Answer:

Let, both blocks start moving with velocity V₁ and V₂ as shown in Figure

Since no horizontal force on the system so, applying momentum

conservation

0 = 40 V₁– 60 V₂

2V₁ = 3V₂ =……..(1)

Now applying energy conservation, Loss in potential energy = gain in kinetic energy

⇒ \(\frac{1}{2} k x^2=\frac{1}{2} m_1 V_1^2+\frac{1}{2} m_2 V_2^2\)

⇒ \(\frac{1}{2} \times 600 \times(1.5)^2=\frac{1}{2} \times 40 \times V_1{ }^2+\frac{1}{2} \times 60 \times V_2{ }^2\) …….

Solving the equation and we get, V1= 4.5 m/s, V2= 3 m/s.

Question 9. Find the mass of the rocket as a function of time, if it moves with a constant acceleration a, in the absence of external forces. The gas escaped with a constant velocity u relative to the rocket and its initial mass was m0₁.
Answer:

Using, Fnet = \(V_{\text {rel }}\left(\frac{-d m}{d t}\right)\)

⇒ \(F_{\text {net }}=-u \frac{d m}{d t}\) …….(1)

F_net = ma ……(2)

Solving equation and

⇒ \(\mathrm{ma}=-\mathrm{u} \frac{\mathrm{dm}}{\mathrm{dt}}\)

⇒ \(\int_{m_0}^m \frac{d m}{m}=\int_0^t \frac{-a d t}{u}\)

⇒ \(\ln \frac{m}{m_0}=\frac{-a t}{u}\)

⇒ \(\frac{m}{m_0}=e^{-a t / u}\)

⇒ \(\mathrm{m}=\mathrm{m}_0 \mathrm{e}^{-\frac{a t}{u}}\)

Question 10. A ball is approaching the ground with speed u. If the coefficient of restitution is e then find out:

1. The velocity just after a collision.
2. The impulse exerted by the normal is due to the ground on the ball.

Answer:

e = \(\frac{\text { velocity of separation }}{\text { velocity of approach }}\)

⇒ \(\frac{v}{u}\)

velocity after collision = V = eu ……..(1)

Impulse exerted by the normal due to ground on the ball = change in momentum of the ball. = {final momentum} – {initial momentum}

= {m v} – {– mu}

= mv + mu = m {u + eu} = mu {1 + e}