Question

A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is *

MV

1.5MV

2MV

Zero​

Answer 1

Answer:

2mv

Explanation:

Let initially the body is moving along positive x- axis with speed v.

So, momentum of the body initially

p

i

=mv

i

^

After bouncing back, the body moves along negative x-axis with same speed.

So, momentum of the body finally

p

f

=−mv

i

^

Thus change in momentum of body Δ

p

=

p

f

−

p

i

=−2mv

i

^

Impulse experienced

I

=Δ

p

=−2mv

i

^

⟹ ∣

I

∣=2mv