Question

1. a ball hits a wall and bounces back at half the original speed. what part of the original kinetic energy of the ball did it lose in the collision?

a. 1/4
b. 1/2
c. 3/4
d. it did not lose kinetic energy.

Explanation:

Step1: Recall Kinetic Energy Formula

Kinetic energy is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is speed. Let the original speed be \( v_1 = v \), so original KE (\( KE_1 \)) is \( \frac{1}{2}mv^2 \).

Step2: Find Final Kinetic Energy

After collision, speed is \( v_2=\frac{v}{2} \). Final KE (\( KE_2 \)) is \( \frac{1}{2}m(\frac{v}{2})^2=\frac{1}{2}m\frac{v^2}{4}=\frac{1}{8}mv^2 \)? Wait, no—wait, the problem says "bounces back at half the original speed"—wait, maybe I misread. Wait, the problem says "bounces back at half the original speed"—so final speed \( v_f=\frac{v_i}{2} \). Then original KE: \( KE_i = \frac{1}{2}mv_i^2 \). Final KE: \( KE_f=\frac{1}{2}m(\frac{v_i}{2})^2=\frac{1}{2}m\frac{v_i^2}{4}=\frac{1}{8}mv_i^2 \)? No, that can't be. Wait, no—wait, maybe the problem is "bounces back at half the original speed"—but maybe the original kinetic energy is considered, and the final kinetic energy is half? Wait, no, the question is: "What part of the original kinetic energy of the ball did it lose in the collision?" Wait, let's re-express. Let original KE be \( KE_1 = \frac{1}{2}mv^2 \). Final speed is \( v' = \frac{v}{2} \), so final KE \( KE_2 = \frac{1}{2}m(v')^2 = \frac{1}{2}m(\frac{v}{2})^2 = \frac{1}{8}mv^2 \)? No, that's not right. Wait, maybe the problem is that the final speed is half, so final KE is \( \frac{1}{2}m(\frac{v}{2})^2 = \frac{1}{4} \times \frac{1}{2}mv^2 = \frac{1}{4}KE_1 \). So the final KE is \( \frac{1}{4} \) of original. Therefore, the energy lost is \( KE_1 - KE_2 = KE_1 - \frac{1}{4}KE_1 = \frac{3}{4}KE_1 \). So the part lost is \( \frac{3}{4} \) of original KE.

Step3: Calculate Lost Fraction

Lost energy \( \Delta KE = KE_1 - KE_2 \). \( KE_2 = \frac{1}{2}m(\frac{v}{2})^2 = \frac{1}{4} \times \frac{1}{2}mv^2 = \frac{1}{4}KE_1 \). Thus, \( \Delta KE = KE_1 - \frac{1}{4}KE_1 = \frac{3}{4}KE_1 \). So the fraction lost is \( \frac{3}{4} \).

Answer:

c. 3/4