Question

A FORCE 10 N ACTS ON A BODY OF MASS 2 KG FOR 3 S,INITIALLY AT REST . CALCULATE (i) velocity acquired by the body (ii) change in momentum of the body.

Answer 1

Explanation:

Force = mass × acceleration

F=10N

m=2kg

then,

a=mF=210=5m/s2

(i) Now, 

u=0

t=3sec

v=u+at=0+3×5=15m/s

(ii)Initialmomentum=mu=2×0=0

       Finalmomentum=mv=2×15=30kg−m/s

Change in momentum=Final momentum - Initial momentum 

                                    =30−0=30kg−m/s

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Answer 2

Given:

• Force = 10N
• Mass = 2 kg
• Time = 3 s

To find:

• i) Velocity acquired by the body
• ii) Change in momentum

Solution :

i) sol:-

Firstly finding Velocity acquired by the body

Using Newton's 2nd law

F = m × a

Where

• F : force
• m : mass
• a : acceleration

→ 10 = 2(a)

→ 10/2 = a

→ acceleration = 5 m/s²

Now , using first equation of motion

v = u + at

Where

• v : final velocity
• u : initial velocity
• a : acceleration
• t : time

Here initial velocity ( u ) will be zero because the body starts from rest

→ v = 0 + 5(3)

→ v = 0 + 15

→ Velocity = 15 m/s

_____________________________

ii) sol:-

Now , finding change in momentum

As we know that

∆p = final momentum - initial momentum

Initial Momentum = m × u

Final momentum = m × v

→ ∆p = mv - mu

→ ∆p = 2(15) - 2(0)

→ ∆p = 30 - 0

→ Change in momentum = 30 kgm/s