A tennis ball of mass = 0.068kg and speed = 21m/s strikes a wall at a 45 degree

Question

A tennis ball of mass = 0.068kg and speed = 21m/s  strikes a wall at a 45 degree angle and rebounds with the same speed at 45 degrees.

What is the magnitude of the impulse given to the ball?

What is the direction of the impulse given to the ball?

Explanation / Answer

let x-axis be horizontal & y-axis be vertically upwards. impulse = change in momentum initial momentum = [0.068*21*sin45] i + [0.068*21*cos45] j => Pi = 1.0097 i + 1.0097 j final momentum = [0.068*21*sin45] (-i) + [0.068*21*cos45] j => Pf = -1.0097 i + 1.0097 j impulse = Pf - Pi => J = (-1.0097 i + 1.0097 j) - (1.0097 i + 1.0097 j) => J = -2.0194 i + 0 j => J = -2.0194 i impulse = 2.0194 N (towards left)

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