Question

28. A particle of mass m and velocity v is divided into two parts of mass m1, and m2 where m1 = 2 × m2. The two parts move in opposite directions to each other. Calculate the speed of the second part when it is known that for the first part it is V1 = 2V. The first part (m1) continues to move in the initial direction, ie “forward.” In what direction will the second part move?

Answer 1

Since $m_1 = 2 m_2$, $m = m_1 + m_2 = 3 m_2$, hence $m_2 = \frac{m}{3}$, $m_1 = \frac{2 m}{3}$.

Let us use the law of conservation of momentum:

$m v = m_1 v_1 - (2 m_1) v_2$, from where the velocity of the second part is:$v_2 = \frac{m_1 v_1 - m v}{2 m_1} = \frac{3(\frac{2}{3} m \cdot 2 v - m v)}{4 m} = \frac{3 m v}{4 m} = \frac{3}{4}v$.

The second part moves in "back" direction.