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/s².

Or, acceleration, a = 4 m/s².

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/s² 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

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 The force is then withdrawn and the body moves with a velocity of 2 m s^-1.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 = (v²  u²)/2s.

Or, a = (0² 100²)/ 2(0.02).

Or, a = 2.5  10⁵ ms^-2.

Therefore, retardation is 2.5  10⁵ ms^-2.

(iv) Force, F = ma

Or, F = (0.05 kg) (2.5  10⁵ ms^-2)

Or, F = 12500 N

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

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