A body weighing 1 kg moves uniformly around a circle at a speed of 2 m / s.

A body weighing 1 kg moves uniformly around a circle at a speed of 2 m / s. Find the change in momentum of the body after it has passed a quarter circle.

m = 1 kg.

V = 2 m / s.

Δp -?

The momentum of a body p is called a vector physical quantity equal to the product of the body’s mass m by its velocity V: p = m * V. The momentum of the body p is directed in the same way as the speed of movement V.

When moving in a circle, the speed V is always directed tangentially to the trajectory of movement, that is, perpendicular to the radius drawn to the point of movement.

Δp = p2 – p1 = m * V2 – m * V1 = m * ΔV.

The body’s velocity V2 through a quarter of a circle will be perpendicular to the initial velocity V1.

The change in speed ΔV will be the hypotenuse of a right-angled triangle with legs V.

By the Pythagorean theorem: ΔV = √ (V1 ^ 2 + V2 ^ 2) = √ (V ^ 2 + V ^ 2) = V * √2.

Δp = m * V * √2.

Δp = 1 kg * 2 m / s * √2 = 2.83 kg * m / s.

Answer: the change in the momentum of the body is Δp = 2.83 kg * m / s.