In the overhead view of the figure a 200 g ball with a speed

In the overhead view of the figure, a 200 g ball with a speed v of 7.0 m/s strikes a wall at an angle of 30degree and then rebounds with the same speed and angle. It is in contact with the wall for 9.0ms. What is the impulse on the ball (take positive as down in the figure)? What is the average force exerted by the ball on the wall (take positive as down in the figure)?

Solution

v_i = v cos theta i - v sin theta j = 7 cos 30 i - 7 sin 30 j = 6.1 i + 3.5 j

v_f = v cos theta i + v sin theta j = 7 cos 30 i + 7 sin 30 j = 6.1 i + 3.5 j

With m = 0.200 kg, the impulse-momentum theorem yields

J = m v_f- m v_i = 2 * 0.200 * 3.5 j

The magnitude of the impulse delivered on the ball by the wall = 1.4 N·s

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The force on the ball by the wall is = J / delta T = 1.4 j / 0.009

= 155.6 N

direction is directly into the wall, or down