## Force And Laws Of Motion Long Answer Type Question And Answers

**Question 1. A cricket ball weighing 200 g moving with a speed of 90 kmhr ^{-1} is caught by a player. While doing so, he moves his hands back by 10 cm. How much force is exerted by**

**the player on the ball?****the ball on the player?**

**Answer.**

Given m = 200 g = 0.2 kg

u = 90 kmhr−1 = \(90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\)

v = 0

S = 10 cm = 0.1 m

F = ?

We know that

v^{2} = u^{2} + 2aS

⇒ 0 = (25)^{2} + 2 × a × 0.1

−0.2 a = 625

⇒ a = \(-\frac{625}{0.2}\)

⇒ a = −3125 ms^{-2}

According to Newton’s II law of motion,

F = ma

⇒ F = 0.2 × −3125

⇒ F = −625 N

Negative sign of force implies that it is a retarding force and acts in a direction opposite to the motion of the ball.

∴ Force exerted by the player on the ball = 625 N

According to Newton’s III law of motion, the ball exerts an equal and opposite force on the players hands

∴ F = 625 N

**Question 2. When a car weighing 800 kg was moving on a horizontal road with a speed of 30 ms ^{-1}, its brakes stopped working. The car came to rest after travelling a distance of 150 m. Find**

**the retardation of the car****frictional force exerted by the ground on the car**

**Answer.**

(1) Given u = 30 ms^{-1}

v = 0

S = 150 m

a = ?

We know that

v^{2}= u^{2} + 2aS

⇒ 0 = (30)^{2} + 2 × a × 150

∴ −300 a = 900

∴ a = −3 ms^{-2}

(2) We know that

F = ma

= 800 × −3

F = −2400 N

Negative sign implies that it is a retarding force. This force which opposes the motion of a car on the ground is a frictional force.

∴ Frictional force exerted by ground on car = 2400 N

**Question 3. A hockey player hits a ball (m = 100 g) lying on ground, the with his stick. It is found that the ball starts moving with a velocity of 40 ms−1.Find**

**the impulse of the force exerted by the stick on the ball.****If the stick was in contact with the ball for 1 ms, then find the magnitude of the force acting on the ball.**

**Answer.**

(1) Impulse of a force = F × t

= ma × t

= \(m \frac{(v-u)}{t} \times t\)

= m(v − u)

= 0.1 kg (40 ms^{-1} − 0)

= 4 kg ms^{-1}

(2) I = F × t

4 kgms^{-1} = F × 1 × 10^{-3} s

∴ F = \(\frac{4 \mathrm{kgms}^{-1}}{1 \times 10^{-3} \mathrm{~s}}\)

= 4 × 10^{3} kgms^{-2}

F = 4 × 10^{3} N

**Question 4. Two trolleys m _{1} = 0.2 kg and m_{2} = 0.3 kg are held against a compressed spring on a smooth horizontal table, as shown in the figure below.**

**When the spring is cut, m _{1} is found to move towards right with an initial acceleration of 6 ms^{-2}. Find the initial acceleration of m_{2}.**

**Answer.**

Force acting on m_{2} is equal and opposite to that force acting on m_{1}.

∴ F_{2} = − F_{1}

i.e., m_{2}a_{2} = − m_{1}a_{1}

0.3 kg × a_{2} = − 0.2 kg × 6 ms^{-2}

∴ a_{2 }= \(-\frac{0.2 \times 6}{0.3}\)

∴ a_{2} = −4 ms^{-2}

Negative sign implies that it is in opposite direction of that of m_{1} i.e., a1.

∴ Acceleration of m_{2} = 4 ms^{-2} towards left.

**Question 5. A toy rocket weighing 500 g is standing vertically on ground. How much force should act on it so that it starts ascending with a uniform acceleration of 5 ms ^{-2}? (Take g = 10 ms^{-2})**

**Answer.**

To go up, the upward force acting on the rocket must be greater than its weight ‘mg’.

Net force acting on rocket

F_{net} = F_{up} − mg (1)

By Newton’s II law,

F_{net} = ma (2)

From equations (1) and (2),

F_{up} − mg = ma

∴ F_{up} = mg + ma

= m (g + a)

= 0.5 kg (10 + 5)ms^{-2}

= 0.5 × 15 kg ms^{-2}

= 7.5 N

**Question 6. Name the physical quantity, which is equal to the rate of change of momentum and derive a relation between them. A car weighing 2000kg moving with a velocity of 90 km/h retards uniformly and comes to rest in 20 s. Calculate the force required to stop the car.**

**Answer.**

Force is the physical quantity, which is equal to rate of change of momentum. Consider an object of mass m moving with an initial velocity u. When a force is acting on the object for a certain time t its velocity changes to v. According to Newton’s second law of motion,

F = ma (1)

Where a is the acceleration produced on the object due to the application of force. Acceleration = Change in velocity/Time

a = v – u/t

Substituting the value of a in equation (1) we get,

F = m (v – u/t)

F = (mv – mu)/t

Since v as is the final velocity, mv will be the final momentum and u is the initial velocity, mu will be the initial momentum.

F = Final momentum – Initial momentum/Time

Force = Rate of change of momentum

Mass of the car = 2000 kg

Initial velocity (u) = 90 km/h

= \(90 \times \frac{5}{18}\)

Initial velocity (u) = 25 m/s

Final velocity (v) = 0

Time (t) = 20 s

Force required to stop the car = m (v – u)/t

= 2000 (0 – 25)/20

= \(-2000 \times \frac{25}{20}\)

= –100 × 25

Force required to stop the car = –2500 N.

**Question 7. A research balloon of total mass M is descending vertically with downward acceleration a as shown in the below figure. How much ballast must be thrown from the car to give the balloon an upward acceleration a, assuming that the upward lift of the air on the balloon does not change?**

**Answer.**

Initially, let us consider the mass of the system (balloon) as M.

So the force ( f ) acting on the system having downward acceleration, a will be,

f = –Ma (1)

And the weight of the system (W) will be,

W = Mg (2)

Where g is the free fall acceleration of the system.

Therefore, the upward force acting on the system (F) will be,

F = W + f = Mg + (–Ma)

F = Mg – Ma (3)

Again let us consider m as the mass of single ballast which is thrown from the balloon.

Now mass of the system is, M – m.

So the force ( f1) acting on the system having an upward acceleration a and mass M – m will be,

f_{1} = (M – m) a (4)

And the weight of the system (W1) having mass M – m will be,

W_{1} = (M – m) g (5)

Where, g is the free fall acceleration of the system.

To give the balloon an upward acceleration a, the upward force (F) must be equal to the addition of the force (f1) acting on the system having upward acceleration a and mass, M-m and the weight of the system (W1) having mass M – m.

So the equation will be,

f_{1} + W_{1} = F (6)

(M – m) a + (M – m) g = Mg – Ma

Ma – ma + Mg – mg = Mg – Ma

ma + mg = Ma + Mg – Mg + Ma

m (a + g) = 2Ma

m = 2Ma/(a + g) (7)

From equation (7) we observed that, 2Ma/(a + g) mass of ballast must be thrown from the carto give the balloon an upward acceleration a.

**Question 8. A meteor of mass 0.25 kg is falling vertically through Earth’s atmosphere with an acceleration of 9.2 m/s2. In addition to gravity, a vertical retarding force (due to frictional drag of the atmosphere) acts on the meteor as shown in the below figure. What is the magnitude of this retarding force?Answer.**

Mass of the meteor, m = 0.25 kg

Acceleration of the meteor, a = 9.2 m/s^{2}

The net force exerted (F_{net}) on the meteor will be,

F_{net} = ma

= (0.25 kg) (9.2 m/s^{2})

= (2.30 kg. m/s^{2}) (1 N/1 kg. m/s^{2})

= 2.30 N (1)

If g (g = 9.80 m/s^{2}) is the free fall acceleration of meteor, then the weight of the meteor (W) will be,

W = mg = (0.25 kg) (9.80 m/s^{2})

= (2.45 kg. m/s^{2}) (1 N/ 1 kg. m/s^{2})

= 2.45 N (2)

The vertical retarding force would be equal to the net force exerted on the meteor (Fnet) minus weight of the meteor (W).

So, vertical retarding force = F_{net} – W (3)

Substituting the value of F_{net} and W in equation (3), the vertical retarding force will be,

Vertical retarding force = F_{net} – W

= 2.30 N – 2.45 N = –0.15 N (4)

From equation (4) we observed that, magnitude of the vertical retarding force would be, –0.15N.

**Question 9. Workers are loading equipment into a freight elevator at the top floor of a building. However, they overloaded the elevator and the worn cable snaps. The mass of the loaded elevator at the time of the accident is 1600 kg. As the elevator falls, the guide rails exert a constant retarding force of 3700 N on the elevator. At what speed does the elevator hit the bottom of the shaft 72 m below?**

**Answer.**

To obtain the weight of the elevator W, substitute 1600 kg for mass of the elevator m and 9.81 m/s^{2} for free fall acceleration g in the equation W = (m) (g).

W = (m) (g)

= (1600 kg) (9.81 m/s^{2})

= 15680 kg, m/s^{2}

= (15680 kg, m/s^{2})(1 N/1 kg ⋅ m/s^{2})

= 15680 N

The magnitude of the net force F will be,

F = W – R

To obtain the magnitude of the net force F, substitute 15680 N for W and 3700 N for retarding force R in the equation F = W – R,

F = W – R

= (15680 N) – (3700 N)

= 11980 N

Rounding off to two significant figures, the magnitude of the net force F will be 12000 N.

To obtain the acceleration a, substitute 12000 N for F and 1600 kg for m in the equation a = F/m, we get,

a = F/m

= 12000 N/1600 kg

= (7.5 N/kg) (1 kg. m/s^{2}/1 N)

= 7.5 m/s^{2}

To obtain the time t to fall, substitute –72 m for y and –7.5 m/s^{2} for a in the equation t = \(\sqrt{2} y / a\)

t = \(\sqrt{2} y / a\)

= \(\sqrt{2}(-72 \mathrm{~m}) /\left(-7.5 \mathrm{~m} / \mathrm{s}^2\right)\)

= 14 s

To obtain the final speed v at which the elevator hits the bottom of the shaft 72 m below, substitute 7.5 m/s^{2} (only magnitude of a) for a and 4.4 s for t in the equation v = at, we get,

v = at

= (7.5 m/s2)(14 s)

= 105 m/s

From the above observation we conclude that, the speed at which the elevator hits the bottom of the shaft 72 m below would be 105 m/s.

**Question 10. A force produces an acceleration of 2.0 m s ^{-2} in a body A and 5.0 m s^{-2} in another body B. Find the ratio of the masses of A to the mass of B.**

**Answer.**

Let the mass of the body A be mAand that of the body B be mB. We have

F = (m_{A}) (2.0 m s^{-2})

F = (m_{B}) (5.0 m s^{-2})

Thus

(mA)(2.0 m s^{-2}) = (m_{B}) (5.0 m s^{-2})

Or \(\frac{m_A}{m_B}=\frac{5.0}{2.0}=2.5\)

**Question 11. A man weighing 60 kg runs along the rails with the velocity of 18 km hr ^{-1} and jumps into a car of mass 1 quintal standing on the rails. Calculate the velocity with which the car will start travelling along the rails.**

**Answer.**

Here, mass of man, m_{1} = 60 kg.

Initial velocity of man,

u_{1}= 18 km/1 h = 18 × 1000 m/60 × 60 s = 5 m s^{-1}.

Mass of car, m_{2} = 1 quintal = 100 kg.

Initial velocity of car, u_{2} = 0.

After a man jumps into a car, let their common velocity be v

Applying the principle of conservation of momentum

Total momentum after jump = Total momentum before jump

(m_{1}+ m_{2}) v = m_{1}u_{1} + m_{2}u_{2}

(60 + 100) v = 60 × 5 + 100 × 0

Or v = \(\frac{300}{160}=1.88 \mathrm{~m} \mathrm{~s}^{-1}\)

**Question 12. A77-kg person is parachuting and experiencing a downward acceleration of 2.5 m/s ^{1} shortly after opening the parachute. The mass of the parachute is 5.2 kg. (1) Find the upward force exerted on the parachute by the air. (2) Calculate the downward force exerted by the person on the parachute.**

**Answer.**

(1) The net force F_{net} on the system is equal to the sum of force exerted by the person and force exerted by the parachute.

So, F_{net} = (m_{pe}+ m_{pa}) (a)

Here, m_{pe} is the mass of person, m_{pa} is the mass of parachute and a is the downward acceleration.

To obtain the net force F_{net} on the system, substitute 77 kg for m_{pe} , 5.2 kg for m_{pa} and –2.5 m/s^{2} for a in the equation F_{net} = (m_{pe} +m_{pa}) (a),

F_{net} = (m_{pe}+ m_{pa}) (a)

= (77 kg + 5.2 kg) (–2.5 m/s^{2})

= (–210 kg,m/s^{2}) (1 N/1 kg,m/s^{2})

= –210 N

The weight W of the system will be,

W = (m_{pe}+ m_{pa}) (g)

To obtain the weight W of the system, substitute 77 kg for m_{pe}, 5.2 kg for m_{pa} and 9.81 m/s^{2} for free fall acceleration g in the equation W = (m_{pe}+ m_{pa}) (g),

W = (m_{pe}+ m_{pa}) (g)

= (77 kg + 5.2 kg) (9.81 m/s^{2})

= (810 kg, m/s^{2}) (1 N/1 kg, m/s^{2})

= 810 N

If P is the upward force of the air on the system (parachute) then,

P = F_{net}+ W = (–210 N) + (810 N) = 600 N

From the above observation we conclude that, the upward force exerted on the parachute by the air would be 600 N.

(2) The net force F_{net} on the parachute will be

F_{net} = m_{pa} a

Here, m_{pa} is the mass of parachute and a is the downward acceleration.

To obtain the net force Fnet on the parachute, substitute 5.2 kg for mpa and –2.5 m/s^{2} for a in the equation F_{net} = mpa a,

F_{net} = m_{pa} a = (5.2 kg) (–2.5 m/s^{2})

= (–13 kg m/s^{2}) (1 N/1 kg m/s^{2}) = –13 N

The weight W of the parachute will be,

W = (m_{pa}) (g)

To obtain the weight W of the system, substitute 5.2 kg for mpa and 9.81 m/s^{2} for free fall acceleration g in the equation W = (m_{pa}) (g),

W = (m_{pa})(g)

= (5.2 kg) (9.81 m/s^{2})

= (51 kg, m/s^{2})(1 N/1 kg, m/s^{2})

= 51 N

If D is the downward force of the person on the parachute then,

D = –F_{net} – W + P

= –(–13 N) – (51 N) + (600 N) = 560 N

**Question 13. Are search balloon of total mass Misdescending vertically with downward acceleration a as shown in below figure. How much ballast must be thrown from the car to give the balloon an upward acceleration a, assuming that the upward lift of the air on the balloon does not change?**

**Answer.**

Let us consider initially mass of the system (balloon) be, M.

So the force ( f ) acting on the system having downward acceleration, a will be,

f = –Ma (1)

And the weight of the system (W) will be,

W = Mg (2)

Where, g is the free fall acceleration of the system.

Therefore the upward force acting on the system (F) will be,

F = W + f = Mg + (–Ma)

F = Mg – Ma (3)

Again let us consider m as the mass of single ballast which is thrown from the balloon.

Now mass of the system is, M-m.

So the force ( f_{1}) acting on the system having upward acceleration, a and mass, M-m will be,

f_{1} = (M – m) a (4)

And the weight of the system (W1) having mass M-m will be,

W_{1} = (M-m) g (5)

Where, g is the free fall acceleration of the system.

To give the balloon an upward acceleration a, the upward force (F) must be equal to the addition of the force (f_{1}) acting on the system having upward acceleration, a and mass, M-m and the weight of the system (W_{1}) having mass M-m.

So the equation will be,

f_{1} + W_{1} = F (6)

(M – m) a + (M – m) g = Mg – Ma

Ma – ma +Mg – mg = Mg – Ma

ma + mg = Ma + Mg – Mg + Ma

m (a + g) = 2Ma

m = 2Ma/(a + g) (7)

From equation (7) we observed that, 2Ma/(a + g) mass of ballast must be thrown from the carto give the balloon an upward acceleration a.

**Question 14. A child’s toy consists of three cars that are pulled in tandem on small frictionless rollers as shown in below figure. The cars have masses m _{1} = 3.1 kg, m_{2} = 2.4 kg, and m3 = 1.2 kg. If they are pulled to the right with a horizontal force P = 6.5 N, find**

**the acceleration of the system,****the force exerted by the second car on the third car, and****the force exerted by the first car on the second car.**

**Answer.**

Given Data:

Mass of the first car, m_{1} = 3.1 kg

Mass of the second car, m_{2} = 2.4 kg

Mass of the third car, m_{3} = 1.2 kg

Horizontal force on the car, P = 6.5 N

Acceleration of a body is equal to the force exerted on the body divided by mass of the body.

(1) The acceleration of the system is equal to the total horizontal force acting on the system divided by total mass of the system.

So,

a = F/ (m_{1}+ m_{2}+ m_{2})

= (6.5 N)/(3.1 kg + 2.4 kg + 1.2 kg)

= (6.5 N)/(6.7 kg)

= (0.97 N/kg)(1 kg. m/s^{2}/1N)

= 0.97 m/s^{2}

Therefore, the acceleration of the system would be 0.97 m/s^{2}.

(2) Force exerted by the second car on the third car would be,

F_{32} = (Mass of third car)(Total acceleration of the system)

= m_{3} a

= (1.2 kg)(0.97 m/s^{2})

= (1.164 kg. m/s^{2})(1 N/1 kg. m/s^{2})

= 1.164 N

Rounding off to two significant figures, the force exerted by the second car on the third car would be 1.2 N.

(3) Force exerted by the first car on the second car would be,

F_{32} = (Sum of mass of second car and third car)(Total acceleration of the system)

= (m_{2}+ m_{3}) a

= (2.4 kg +1.2 kg)(0.97 m/s^{2})

= (3.492 kg. m/s^{2})(1 N/1 kg. m/s^{2})

= 3.492 N

Rounding off to two significant figures, the force exerted by the first car on the second car would be 3.5 N.