Question

A man pushes a 500 kg car initially at rest, with a constant force of 500
Newtons for a time of 10 seconds. Ignore any friction. what is the change in
momentum of the car?​

Answer 1

GiveN :-

• A man pushes a 500 kg car initially at rest, with a constant force of 500 Newtons for a time of 10 seconds.
• Ignore any friction.

To FinD :-

• The change in momentum of the car.

SolutioN :-

Given that a man pushes a 500 kg car initially at rest, with a constant force of 500 Newtons for a time of 10 seconds. We need to find the change in momentum . Also we know that the rate of change of Momentum is called Force. Let the change in Momentum represented by ∆p . So , let's substitute upon the respective values.

$\sf\dashrightarrow \pink{ Force (F)=\dfrac{\Delta p }{time } }\\\\\sf\dashrightarrow \dfrac{\Delta p }{10s }= 500 N \\\\\sf\dashrightarrow \Delta p = 500 N \times 10s \\\\\sf\dashrightarrow \underset{\blue{\sf Required \ change }}{\underbrace{\boxed{\pink{\frak{\Delta p = 5000 kg-m/s }}}}}$

Answer 2

Answer:

5000 kg m/s

Explanation:

Let the final velocity be 'v',

Initial velocity = u = 0(initially, at rest)

Final velocity = v

Using, v = u + at ,after 10 seconds,

=> v = 0 + a(10)

=> v/10 = a ...(1)

Given,

=> F = 500 N

=> ma = 500 => 500a = 500

=> a = 1

From (1), a = v/10,

=> v/10 = 1 => v = 10

Thus, final velocity of the car is 10 m/s

Final momentum = mv = 500 kg x 10 m/s

Initial momentum = mu = 500 kg x 0

Change in momen. = 5000 kg m/s - 0

∆p(momentum) = 5000 kg m/s

This is related to your previous question.

Alternatively, for straight line motion,

F = ma

F = m(v - u)/t {v - u = at}

F = (mv - mu)/t

F = (∆p)/t

500 = ∆p/10

5000 = ∆p