Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 9 - Center of Mass and Linear Momentum - Problems - Page 251: 53a

Answer

The percentage of the original kinetic energy that was lost in the collision is $~~33.3\%$

Work Step by Step

Let $v_0$ be the speed of the car before the collision. We can find an expression for the initial kinetic energy: $K_i = \frac{1}{2}mv_0^2$ $K_i = \frac{1}{2}(1000)v_0^2$ $K_i = 500~v_0^2$ We can use conservation of momentum to find an expression for the speed after the collision: $p_f = p_i$ $(1500~kg)~v_f = (1000~kg)~v_0$ $v_f = \frac{(1000~kg)~v_0}{1500~kg}$ $v_f = \frac{2}{3}~v_0$ We can find an expression for the final kinetic energy: $K_f = \frac{1}{2}mv_f^2$ $K_f = \frac{1}{2}(1500)(\frac{2}{3}~v_0)^2$ $K_f = \frac{1000}{3}~v_0^2$ We can find the percentage of the original kinetic energy that was lost: $\frac{K_i-K_f}{K_i}\times 100\% =\frac{500~v_0^2-\frac{1000}{3}~v_0^2}{500~v_0^2}\times 100\% = 33.3\%$ The percentage of the original kinetic energy that was lost in the collision is $~~33.3\%$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.