Question

if a particle of mass 'm' strikes a wall with a velocity 'v' at an angle of 45 degree elastically, the impulse imparted to the wall is :

a)mv/2 b)mv c)square root of 2 multiply by mv d)2mv

Answer 1

If a particle of mass m strikes a wall with a velocity v at an angle 45° elastically with horizontal line after striking particle is moving with same magnitude of velocity v and direction of velocity is shown in figure.
Here impulse is along to horizontal component of velocities.

So, horizontal component of velocity of particle just before striking is vcos45° [ right direction ]
horizontal component of velocity of particle just after striking is -vcos45° [ negative sign indicates direction is in left]

Now, impulse = change in momentum
= mvco45° - m(-vcos45°)
= 2mvcos45°
= 2mv × 1/√2 = √2mv ,

Hence, impulse = √2mv , answer should be option (C)

Answer 2

Correct Answer is C: √2mv 
Explanation:

Momentum is defined as product of mass and velocity.

Linear momentum of particle before  striking=mv Cos 45

momentum of particle  after striking = -mvcos45

Impulse= Change in linear momentum

=mvcos45-(mvcos45)

=2mvcos45

=2mvx1/√2
=√2mv
Hence impulse imparted to wall =√2mv