Question

A body of mass 6kg travelling with a velocity 10 m/s collides head on and elastically with a body of mass 4kg travelling at a speed 5 m/s in opposite direction. The velocity of the second body after the collision is.

Answer 1

According to the momentum conservation law

$$P = m _ {1} V _ {1} - m _ {2} V _ {2} = m _ {2} V _ {4} - m _ {1} V _ {3} = 6 * 10 - 4 * 5 = 40 \frac {\mathrm {kg\,m}}{\mathrm {s}}$$

According to the energy conservation law

$$K = \frac {m _ {1} V _ {1} ^ {2}}{2} + \frac {m _ {2} V _ {2} ^ {2}}{2} = \frac {m _ {1} V _ {3} ^ {2}}{2} + \frac {m _ {2} V _ {4} ^ {2}}{2} = 6 * \frac {10 ^ {2}}{2} + 4 * \frac {5 ^ {2}}{2} = 350\,J$$

Hence

$$V _ {4} = 13 \frac {m}{s}$$