Problem 11 - College Physics

Question

A ball of mass \(0.150 \mathrm{~kg}\) is dropped from rest from a height of \(1.25 \mathrm{~m}\). It rebounds from the floor to reach a height of \(0.960 \mathrm{~m}\). What impulse was given to the ball by the floor?

Step-by-Step Solution

Answer

Let's input the given values into the equations described in the steps above. The initial and final velocities are calculated to be approximately \(v_i = 4.97 \, m/s\) and \(v_f = 4.37 \, m/s\). The impulse imparted by the floor is calculated as approximately 1.40 kg.m/s.

Step 1: Calculate Initial and Final Velocities

Firstly, consider the ball's initial velocity before hitting the floor which is calculated using the kinematic equation. The velocity of an object under free fall is given by \(v = \sqrt{2gh}\), where \(g\) is the acceleration due to gravity (9.81 m/s²) and \(h\) is the height (1.25 m). So, \(v_i = \sqrt{2 * 9.81 * 1.25}\). After bouncing back, the ball achieves a height of 0.96 m, which is used to calculate the final velocity, \(v_f = \sqrt{2 * 9.81 * 0.96}\). These velocities will be used to determine the momentum.

Step 2: Find the change in momentum

The momentum of an object is given by the product of its mass and velocity. Therefore, we calculate the initial momentum before impact as \(p_i = m * v_i\) and the final momentum after impact as \(p_f = m * v_f\). The impulse imparted by the floor results in a change of momentum, given by \(\Delta p = p_f - p_i\). The direction of the velocity (and hence momentum) changes upon the rebound, so we need to consider one of the momenta as negative while calculating the change in momentum. Here the initial momentum will be considered negative as it is downwards, so \(\Delta p = m * v_f - ( - m * v_i ) = m * v_f + m * v_i\).

Step 3: Calculate Impulse

Impulse is defined as the change in momentum. Therefore, the impulse imparted by the floor is equal to the change in momentum. Hence, the last step is to calculate the value from the change in momentum derived in step 2. Thus, the impulse is given by: Impulse = \(\Delta p\)