# Laws of Motion Exe-3C Numericals Answer Physics Class-9 ICSE Selina Publishers

Laws of Motion Exe-3C Linear Momentum Newton’s Second Law Numericals Answer Type for Class-9 ICSE Concise Physics. There is the solutions of Numericals Answer type Questions of your latest textbook which is applicable in 2023-24 academic sessionVisit official Website CISCE for detail information about ICSE Board Class-9.

## Laws of Motion Exe-3C Linear Momentum Newton’s Second Law Numericals Answer

(ICSE Class – 9 Physics Concise Selina Publishers)

 Board ICSE Class 9th Subject Physics Writer / Publication Concise selina Publishers Chapter-3 Laws of Motion Exe – 3C Linear Momentum Newton’s Second Law Topics Solution of Exe-3(C) Numericals Answer Type Academic Session 2023-2024

### Exe-3C Linear Momentum Newton’s Second Law Numericals Answer Type

Ch-3 Laws of Motion Physics Class-9 ICSE Concise

Page 77

#### Question 1. A body of mass 5 kg is moving with velocity 2 m s-1. Calculate its linear momentum.

Mass of the body, m = 5kg

Velocity, v = 2 m/s

Linear momentum = mv = (5)(2) kg m/s

= 10 kg m/s-1

#### Question 2. The linear momentum of a ball of mass 50 g is 0.5 kg m s-1. Find its velocity.

Linear momentum = 0.5 kg m/s

Mass, m = 50 g = 0.05 kg

Velocity = Linear momentum/mass

= 0.5/0.05 m/s

= 10 m/s -1

#### Question 3. A force of 15 N acts on a body of mass 2 kg. Calculate the acceleration produced.

Force, F = 15 N

Mass, m = 2kg

Acceleration, a = F/m [ From Newton’s second law]

Or, a = (15/2) ms-2

Or, a = 7.5 ms-2

#### Question 4. A force of 10 N acts on a body of mass 5 kg. Find the acceleration produced.

Force, F = 10 N

Mass, m = 5kg

Acceleration, a = F/m [ From Newton’s second law]

Or, a = (10/5) ms-2

Or, a = 2 ms-2

#### Question 5. Calculate the magnitude of force which when applied on a body of mass 0.5 kg produces an acceleration of 5 m s-2.

Mass, m = 0.5 kg.

Acceleration, a = 5 ms-2

Force, F = ma   [ From Newton’s second law]

Or, F = (0.5) (5) N = 2.5 N.

#### Question 6. A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. Calculate: (i) The velocity acquired by the body and (ii) Change in momentum of the body.

Force, F = 10 N

Mass, m = 2 kg

Time, t = 3 s

Initial velocity, u = 0 m/s.

(i) Let v be the final velocity acquired.

From Newton’s second law,

F = ma.

Or, a = F/m = 10/2 = 5 ms-2.

From the 1st equation of motion,

a = (vu)/t

Or, v = at + u.

Or, v = (5)(3) + 0 = 15 m/s-1

(ii) Change in momentum = Final momentum  initial momentum

p = mv  mu.

Or, p = m (vu).

Or, p = 2 ( 15 0) = 30 kg m/s-1

(Laws of Motion Exe-3C Numericals Class-9 ICSE)

#### Question 7. A force acts for 10 s on a stationary body of mass 100 kg, after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate: (i) The velocity acquired by the body, (ii) The acceleration produced by the force and (iii) The magnitude of the force.

Mass, m = 100 kg

Distance moved, s = 100 m

Initial velocity, u = 0

(i) Because the body moves through a distance of 100 m in 5 s,

Velocity of the body = Distance moved / time taken

Velocity = (100/5) = 20 m/s-1

(ii) From Newton’s third equation of motion,

v u2 = 2as.

Or, a = (v u2) /2s.

and, a = [ (202  02)/ 2(100) ] ms-2.

hence, a = 2 ms-2.

(iii) Force, F = ma

Or, F = (100) (2) N.

therefore, F = 200 N.

#### 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 vgraph)

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.

Laws of Motion Exe-3C Numericals Class-9 ICSE

### Ch-3 Laws of Motion Physics Class-9 ICSE Concise

Page 78

#### 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?

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. 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.

Force = Rate of change of Momentum
Change in momentum = final momentum – initial momentum = m ( vf – v)
where m = 150 g is mass of ball , vf is final speed and vis initial speed
Change in momentum = 0.150 × ( 25 – 0 ) = 3.75 kg m/s
Rate of change of momentum = change in momentum / time = 3.75 / 0.03 = 125 N
Force applied to stop the ball = 125 N

#### 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.

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 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.

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.

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.

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,

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. 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.

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 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 Exe-3C Linear Momentum Newton’s Second Law Numericals Answer Type  Solutions :–

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