Ball A moves down a frictionless inclined plane. Its velocity v_A at the base of the plane is determined from the law of conservation of energy:

mAg.4h = 1 2 mAvA2

The gravitational potential energy (given on the left side) is transformed into kinetic energy at the base of the plane. Here the zero level for the potential gravitational energy is fixed to be the horizontal at the base of the plane, and the gravitational field of the Earth is supposed to be uniform.

From the previous equation, v_A is determined as:

vA=2

___ Ö2gh

.

After that, ball A collides with ball B at rest. We assume ball A is moving horizontally with velocity v_A when it collides with ball B. After the collision the horizontal velocities of balls A and B are u_A and u_B respectively. The collision is characterised by the coefficient of restitution k:

k = uB - uA vB - vA

and by the law of conservation of momentum:

mAvA + mBvB = mAuA + mBuB

In this particular case, m_A = 2m, m_B = m, v_B = 0, k = 0.5. Solving the system of 2 equations for the unknowns, u_A and u_B: u_B = v_A and u_A = 0.5v_A, i.e. after collision both balls move in the same horizontal direction as the direction of ball A before collision, ball B quicker than ball A.

Ball B is then projected horizontally with velocity v_B. Associating to the movement a system of cartesian axes and measuring time from t=0 when ball B is at height h then when the ball hits the ground:

x = uBt

and

y = 1 2 gt2 = h.

The horizontal distance traveled by ball B is

x = uB   æ  ú Ö 2h g   = vA   æ  ú Ö 2h g   =2   ___ Ö2gh     æ  ú Ö 2h g   =4h.