Question

If a particle of mass m strikes a wall with a velocity v at an angle 45° elastically, the impulse imparted to the wall is

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