Question

A ball with an initial velocity of 10 m/s moves at an angle 60º above the +x-direction. The ball hits a vertical wall and bounces off so that it is moving 60º above the −x-direction with the same speed. In terms of m, the mass of the ball, what is the impulse delivered by the wall?

Answer 1

The impulse delivered by the wall can be calculated using the principle of conservation of momentum. The change in momentum of the ball can be calculated by multiplying the mass of the ball (m) by the change in velocity. The impulse delivered by the wall can be calculated as: Impulse = m * change in velocity.

Step-by-step explanation:

The impulse delivered by the wall can be calculated using the principle of conservation of momentum. When the ball hits the wall, it experiences an impulse in the opposite direction. The change in momentum of the ball can be calculated by multiplying the mass of the ball (m) by the change in velocity.

Given that the ball initially moves at an angle of 60º above the +x-direction and after bouncing off, it moves at an angle of 60º above the -x-direction, we can calculate the change in velocity along the x-axis. The initial velocity along the x-axis is 10 m/s * cos(60º), and the final velocity along the x-axis is 10 m/s * cos(60º). Hence, the change in velocity along the x-axis is 2 * 10 m/s * cos(60º).

Therefore, the impulse delivered by the wall can be calculated as:

Impulse = m * change in velocity = m * 2 * 10 m/s * cos(60º)