A ball of mass 1kg is rolled down a hill of height 10m. At the bottom it collides with another ball of mass 5kg. What speed does the second ball move away with? You can assume the collision between the balls is elastic.

First, we look at the energy of the smaller ball.We know Gravitational potential energy = mgh (m=mass g=9.81 h=height)So we can plug our numbers from the question into the equation to get:GPE=mgh=1x9.81x10=98.1JNow we have the kinetic energy at the bottom of the hill which can be written as:KE=0.5mv^2, which rearranged gives:v=sqrt(2KE/m)=sqrt(298.1/1)=14.01ms^-1So the ball's velocity at the bottom of the hill = 14.01ms^-1Now the second part of the question:We can assume the collision is elastic, so from conservation of momentum we know:m1v1=m2v2, so114.01=5*v (where v is the velocity were looking for), so:v=14.01/5=2.80ms^-1So the 5kg ball moves away with velocity 2.80ms^-1.