How much will the impulse of the body change if a force of 50N acts on it for 0.02 s?

t = 0.02 s.

F = 50 N.

Δp -?

The momentum of a body p is the product of its speed of motion V by its mass m: p = m * V.

Since the body does not change its mass during movement, then Δp = m * ΔV.

Let’s write 2 Newton’s law: F = m * a, where a is the acceleration that the body receives under the action of force.

The acceleration of the body a is expressed by the formula: a = (V – V0) / t = ΔV / t.

2 Newton’s law will take the form: F = m * ΔV / t = Δp / t.

The change in the momentum of the body Δp will be expressed by the formula: Δp = F * t.

Δp = 50 N * 0.02 s = 1 kg * m / s.

The product of the force F by the time of its action t is called the impulse of the force F * t.

The change in the momentum of the body Δp is equal to the impulse of the force F * t, which acts on it – 2 Newton’s law.

Answer: under the action of the force, the momentum of the body will change by Δp = 1 kg * m / s.