Impulse experienced by a body hitting a wall

Explanation:
When a body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity, it experiences an impulse.

Impulse: The change of momentum of an object is called impulse. It is given by the product of force and time taken for the force to act.

Calculation:
The initial momentum of the body = MV
When it hits the wall, it experiences a force F.
The time taken by the force to act is the time taken by the body to come to rest and then move back with the same velocity.
Let T be the time taken for the body to stop and then move back with the same velocity.
Then, the impulse experienced by the body is given by:
∆p = F × T

Now, using the law of conservation of energy, we can find the force experienced by the body.

The initial kinetic energy of the body = (1/2)MV²
The final kinetic energy of the body = (1/2)MV²
The work done by the force is zero, as the displacement is zero.
Therefore, the initial kinetic energy = final kinetic energy
(1/2)MV² = 0 + 0
MV² = 0
V = 0

This means that the body comes to rest when it hits the wall and then moves back with the same velocity.

Now, using the equation of motion, we can find the time taken for the body to stop and then move back with the same velocity.

v = u + at
0 = V + aT
T = -V/a

The acceleration of the body is given by:
a = F/M

Therefore, the time taken for the body to stop and then move back with the same velocity is given by:
T = -MV/F

The impulse experienced by the body is given by:
∆p = F × T
∆p = F × (-MV/F)
∆p = -MV²

Therefore, the impulse experienced by the body is -MV², which is equal to 2MV as the velocity of the body changes from +V to -V. Hence, the correct answer is option 'C' (2MV).