Question

A rock A moving with horizontal speed of 10.0 m/s hits rock B which was at rest, and which has 2 times less mass then A. After the collision, the rocks scatter off, so that A goes 20 degrees above the horizontal, whereas B goes 35 degrees above the horizontal. Find the speeds vfA and vfB that the rocks have just after the collision. Ignore friction, or any other external forces acting in the horizontal plane! Note: In the end, you need to solve tow equations with two unknown speeds!

I know you use the x and y equations with the Initial masses*velocities of A + initial masses*velocities of B = final masses*velocities of A + final masses*velocities of B

I believe the answers should be 7 & 8.3

Answer 1

In the question I think there is mistake for direction given for 2 rock ..it should be 35 degree below the horizontal...I have solved in taking it to be below

Now mass of A = 2m mass of B = m v_1 = 10 m/s v_2 = 0 Before collision By concentration of momentum {Along X - axis} m_1 v_1 + m_2 v_2 = m_1 v_1 cos theta_1 + m_2 v_2 cos theta_2 v_1 and v_2 are final speeds 2m times 10 + 0 2 m v_1 cos 20 degree + m v_2 cos 35 degree 20 m = m (1.8795 v_1) - 0.81932 v_2 1.8795 v_1 + 0.81932 v_2 = 20 rightarrow (1) Along y - axis 0 = m_1 v_1 sin theta_1 - m_2 v_2 sin theta_2 0 = 2 m_1 v_1 sin 20 degree - m v_2 sin 35 degree 0 = m (0.6837 v_1 - 0.57332 v_1) 0.6837 v_! - 0.533 v_2 = 0 rightarrow (2) On solving eq (1) and (2) v_1 = 7 m/s v_2 = 8.3 m/s