Question

x Incorrect. An object, with mass m and speed vi relative to an observer, explodes into two pieces, one 3 times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system during the explosion, as measured in the observer's reference frame? State your answer in terms of the given variables. Click here to enter or edit your answer 4(K. E)-6 (11)

Answer 1

Solution:

Initial kinetic energy of the system

$KE_{i}=\frac{1}{2}mv_{i}^{2}$

Conservation of momentum

$mv_{i}=\frac{3}{4}mv_{f}$

$v_{f}=\frac{4}{3}v_{i}$

Final kinetic energy of the system

$KE_{f}=\frac{1}{2}\left ( \frac{3}{4}m \right )v_{f}^{2}$

4 KEF 1/3 am 3

$KE_{f}=\frac{4}{3}\left ( \frac{1}{2}mv_{i}^{2} \right )$

The kinetic energy added

$\Delta KE=KE_{f}-KE_{i}=\frac{4}{3}\left ( \frac{1}{2}mv_{i}^{2} \right )-\left ( \frac{1}{2}mv_{i}^{2} \right )$

$\Delta KE=\frac{1}{6}mv_{i}^{ 2}$