A body of mass m moves with a speed v. After interacting with the wall, the body began to move

A body of mass m moves with a speed v. After interacting with the wall, the body began to move in the opposite direction with the same velocity in absolute value. What is the modulus of the body’s impulse change?

Given:

m (kilogram) – body weight;

v (meter per second) – body speed.

It is required to determine dp (kg * m / s) – the modulus of the body impulse change after interaction with the wall.

Let us introduce a coordinate system, the OX axis of which is directed towards the initial motion of the body.

Then, the momentum of the body before the interaction will be equal to:

p1 = m * v.

According to the condition of the problem, after interaction with the wall, the velocity of the body remained the same in magnitude, but opposite in direction. After the interaction, the momentum of the body will be equal to:

p2 = m * (-v) = -m * v.

Then the modulus of the impulse change will be equal to:

dp = | dp | = | p2 – p1 | = | -m * v – m * v | = | -2 * m * v | = 2 * m * v.

Answer: the modulus of the pulse change will be equal to 2 * m * v