Question

a body of mass M hits normally a rigid wall with velocity V and bounce back with same velocity. the impulse experienced by the body is??

Answer 1

answer is 3) 2 MV
Impulse=MV-(-MV)=2MV

Answer 2

A body of mass M hits normally a rigid wall with velocity v.
so, initial momentum of body , $P_i=Mv$

body bounces back with same magnitude of velocity. but direction of velocity is just opposite. e.g., $v_f=-v$
so, final momentum of body , $P_f=mv_f$
$P_f=-Mv$

we know, impulse = change in momentum
so, impulse experienced by wall = $P_f-P_i$
impulse experienced by wall = -Mv - Mv = -2Mv
so, impulse experienced by body = - impulse experienced by wall
= -(-2Mv) = 2Mv

hence, option (c) is correct.