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(2*98.1/1)=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, so1*14.01=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.