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(a) MV(b) 1.5MV(c) 2MV(d) zero

Answer 1

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A body of mass M hits normally a rigid wall with velocity v.

so, initial momentum of body pi = mv

body bounces back with same magnitude of velocity. pf=mvf,pf = -mv

but direction of velocity is just opposite. e.g., pv=-v
so, final momentum of body , pf=mvf

we know, impulse = change in momentum
so, impulse experienced by wall = pf-pi

impulse experienced by wall = -Mv - Mv = -2Mv

so, impulse experienced by body = - impulse experienced by wall

$\huge\boxed{\texttt{\fcolorbox{red}{aqua}{ -(-2Mv) = 2Mv}}}$

Answer 2

Answer:

C) 2MV

Explanation:

A body of mass M hits normally a rigid wall with velocity v.

so, initial momentum of body = Pi = mv

The body will bounce back with same magnitude of velocity, but the direction of velocity will be opposite. = vf = -v

Thus, the final momentum of body = Pf = mvf

Pf = -mv

Impulse = change in momentum

Impulse experienced by wall = Pf-Pi

impulse experienced by wall = -Mv - Mv = -2Mv

Impulse experienced by body = - impulse experienced by wall

= (-2Mv)

= 2MV

Thus, the impulse experienced by the body is 2MV