Question

If the momentum of abody increases by 20%,then what is the percentage in its kinetic energy?

Answer 1

Question 15461

We are given, $p_0, p_1 = 1.2p_0$ . One knows, that kinetic energy is given by $T = \frac{mv^2}{2} = \frac{p^2}{2m}$ . Hence, $\frac{T_1}{T_0} = \frac{p_1^2}{p_0^2} = (1.2)^2 = 1.44$ , so the kinetic energy increases by $44\%$ .