# Exercise-3C Laws of Motion Concise ICSE Class-9 Selina Publishers

**Question 8**

Fig 3.11. shows the velocity-time graph of a particle of mass 100 g moving in a straight line. Calculate the force acting on the particle.

(Hint : Acceleration = Slope of the *v*–*t *graph)

#### Answer 8

Slope of a velocity-time graph gives the value of acceleration.

Here, slope = 20/5 = 4 m/s^{2}.

Or, acceleration, a = 4 m/s^{2}.

Force = Mass × Acceleration.

Given mass, m = 100 g = 0.1 kg.

Force = (0.1) (4) = 0.4 N.

**Question 9**

A force causes an acceleration of 10 m s^{-2} in a body of mass 500 g. What acceleration will be caused by the same force in a body of mass 5 kg?

**Answer 9**

Let the force be F.

Force F causes an acceleration, a = 10 m/s^{2} in a body of mass, m = 500 g or 0.5 kg

Thus, F = ma

Or, F = (0.5) (10) = 5 N

Let a’ be the acceleration which force F (=5N) cause on a body of mass, m’ = 5 kg.

Then, a’ = F/m’.

Or, a’ = (5/5) ms^{-2}.

Or, a’ = 1 ms^{-2}.

**Question 10 Exercise-3C Laws of Motion Concise ICSE **

A cricket ball of mass 150 g moving at a speed of 5 ms-1 is brought to rest by a player in 0.003 s. Find the average force applied by the player.

**Answer 10**

_{f}– v

_{i })

_{f}is final speed and v

_{i }is initial speed

**Question 11**

A force acts for 0.1 s on a body of mass 2.0 kg initially at rest. The force is then withdrawn and the body moves with a velocity of 2 m s^{-1}. Find the magnitude of the force.

**Answer 11**

Mass, m = 2 kg

Initial velocity, u = 0

Final velocity, v = 2 m/s

Time, t = 0.1 s

Acceleration = Change in velocity/time

Or, a = (v u) /t

Or, a = (2 0)/ 0.1 = 20 ms^{-2}.

Force = Mass Acceleration

Or, F = (2) (20) = 40 N.

**Question 12**

A body of mass 500 g, initially at rest, is acted upon by a force which causes it to move a distance of 4 m in 2 s, Calculate the force applied.

**Answer 12**

**Question 13**

A car of mass 480 kg moving at a speed of 54 km per hour is stopped in 10 s. Calculate the force applied by the brakes.

**Answer 13**

Mass, m = 480 kg.

Initial velocity, u = 54 km/hr = 15 m/s.

Final velocity, v = 0.

Time, t = 10 s.

Acceleration = Change in velocity/time.

Or, a = (v u)/t.

Or, a = (015)/10 = -1.5 ms^{-2}.

Here, negative sign indicates retardation.

Now, Force = Mass Acceleration

Or, F = (480) (1.5) = 720 N.

**Question 14**

A car is moving with a uniform velocity 30 ms^{-1}. It is stopped in 2 s by applying a force of 1500 N through its brakes.

Calculate the following values:

(a) The change in momentum of car.

(b) The retardation produced in car.

(c) The mass of car.

**Answer 14**

Initial velocity, u = 30 m/s

Final velocity, v = 0

Time, t = 2s

Force, F = 1500 N

Here, a = (v u)/t = (0 30)/ 2 = 15 ms^{-2}. Here, negative sign indicates retardation.

Now, F = ma.

Or, m = F/a = (1500/ 15) = 100 kg.

(a) Change in momentum = Final momentum – Initial momentum

Or, p = m (vu)

and , p = 100 (0 30)

Hence, p = 3000 kg m/s^{-1}

(b) Acceleration, a = (vu)/t.

Or, a = (0 30)/ 2 = 15 ms^{-2},

Here, negative sign indicates retardation.

Thus, retardation = 15 ms^{-2}.

(c) From Newton’s second law of motion,

F = ma

Or, m = F/a = (1500/ 15) = 100 kg.

**Question 15 Exercise-3C Laws of Motion Concise ICSE **

A bullet of mass 50 g moving with an initial velocity 100 m s^{-1} strikes a wooden block and comes to rest after penetrating a distance 2 cm in it.

Calculate:

(i) Initial momentum of the bullet,

(ii) Final momentum of the bullet,

(iii) Retardation caused by the wooden block and

(iv) Resistive force exerted by the wooden block.

**Answer 15**

Mass, m = 50 gm = 0.05 kg.

Initial velocity, u = 100 m/s.

Final velocity, v = 0.

Distance, s = 2cm = 0.02 m.

(i) Initial momentum = mu = (0.05) (100) = 5 kg m/s^{-1}

(ii) Final momentum = mv = (0.05) (0) = 0 kg m/s.

(iii) Acceleration, a = (v^{2 } u^{2})/2s.

Or, a = (0^{2} 100^{2})/ 2(0.02).

Or, a = 2.5 10^{5} ms^{-2}.

Therefore, retardation is 2.5 10^{5} ms^{-2}.

(iv) Force, F = ma

Or, F = (0.05 kg) (2.5 10^{5} ms^{-2})

Or, F = 12500 N

–: End of Laws of Motion Class-9 Physics Selina Solutions :–

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