# The Momentum Questions And Solutions

## The physics lesson, momentum subject. The momentum of an object that going with speed of V. Calculation of the momentum. Relationship between mass, velocity and momentum. Questions and answers.

Question -1

An object has a mass of 10 kg and a velocity of 20 m/s. This object hits a barrier and go back with velocity of 15 m/s.

What is the change in momentum of this object?

Solution:

The change in momentum of the object is found as follows.

ΔP = Plast – Pfirst

The momentum of the body is

P1 = 10•20 = 200 N•s

The momentum of the body is

P2 = -10•15 = -150 N•s

ΔP = P2 – P1

ΔP = -150 – 200

= - 350 N•s

The sign of the result is negative (-), because the direction of the object is reversed. The magnitude of change in the momentum of the object is 350 N•s.

Question -2

A vehicle is travelling at velocity of 25 m/s.

This vehicle increase its velocity to 32 m/s after a while.

Mass of the vehicle is 500 kg.

Find the change in the momentum of the vehicle.

Solution:

First velocity of the vehicle is

V1 = 25 m/s

Last velocity of the vehicle is

V2  = 32 m/s

Mass of the vehicle is

m = 500 kg

First momentum of the vehicle is

P1 = 25•500

P1 = 12500 N•s

Last momentum of the vehicle is

P2 = 32•500

P2 = 16000

Change in the momentum is

ΔP = P2 – P1

ΔP = 16000 – 12500

ΔP = 3500 N•s

Question -3

A car that is motionless is starting travel and its velocity is being 30 m/s after a while.

The mass of the car is 800 kg.

Find the change in the momentum of the car.

Solution:

First velocity of the car is

V1 = 0 m/s

Last velocity of the car is

V2 = 30 m/s

The mass is

m = 800 kg

First momentum of the car is

P1 = 0•800 = 0

Last momentum of the car is

P2 = 30•800 = 24000

The change in the momentum of the car is

ΔP = 24000 – 0

= 24000 N•s

Question -4 The body in the figure is moving toward the barrier at a velocity of 15 m/s. The body is hits the barrier and going back at a velocity of 12 m/s as in the figure .

The mass of the body is 15 kg.

Find the change in the momentum of the body.

(sin37 = 0,6, sin53 = 0,8)

Solution: We must find perpendicular components of the velocities of the object to the barrier . Then, we multiplication these velocities by the mass.

The perpendicular component of the V1 to the barrier is.

V1x = 0.6 V1

V1x = 15•0.6

V1x = 9 m/s

First momentum is

P1 = m•V1x

P1 = 15•9

P1 = 135 N•s

The perpendicular component of the V2 is

V2x = 0.8V2

V2x = 12•0.8

V2x = 9.6 m/s

The last momentum is

P2 = 15•9.6

P2 = -144 N•s

ΔP = -144 – 135

ΔP = -279 N•s

ΔP = 279 N•s

Question -5 The velocity-time graph of an object K is as in the figure.

The mass of the object K is 12 kg.

What is the change in momentum of the object between 0 and 8 seconds?

Solution:

The velocity of the object is 15 m/s instant of t=0 . The velocity of the object was increases to 27 m/s after 27 second. The change in the velocity is 12 m/s.

P1 = 15•12 = 180 N•s

P2 = 27•12 = 324 N•s

ΔP = Plast – Pfirst

ΔP = 324 – 180

ΔP = 144 N•s

RISE KNOWLEDGE

February 14 2019

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