**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 session**. **Visit 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.

**Answer:**

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.

**Answer:**

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.

**Answer:**

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.

**Answer:**

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}.

**Answer:**

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.

**Answer:**

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 1^{st} 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.

**Answer:**

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^{2 } u^{2} = 2as.

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

and, a = [ (20^{2} 0^{2})/ 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 *v*–*t *graph)

**Answer:**

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.

**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?

**Answer:**

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. **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:**

_{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:**

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:**

**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:**

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:**

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. **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:**

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

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