1. The problem statement, all variables and given/known data 2 balls hang from a modified Newtons Cradle, on of mass 80g, the other of mass 200g. Assuming gravity is 9.81ms^-2 ball A is pulled out so it reaches a height of 10cm then released. After impact ball 2 moves at a speed of 0.80ms^-1... calculate the velocity of sphere A immediately after impact. A total of 9 marks 2. Relevant equations E grav = mgh E trans = 0.5mv^2 The law of conservation of momentum 3. The attempt at a solution (a) m1 = 0.08kg m2 = 0.2kg v2 = 0.8ms-1 v1 = ? height = 0.1m Neglecting air resistance when ball A is pulled up to a height of 0.10m where its gravitational potential energy is at a maximum: Egrav = mgh = 0.08kg x 9.81ms-1 x 0.1m = 0.078 Joules The ball is then released and Egrav is transferred to kinetic energy Etrans fully when the ball reaches the bottom of the swing so Egrav will be at a minimum and Etrans will be at a maximum. So if Etrans = ￼0.5mv^2 then assuming there is no energy loss then Etrans = 0.078J Therefore rearranging Etrans for v then the velocity at the point of impact is: v1 ￼= 1.4ms-1 at the point of impact. The kinetic energy of m2 after being struck is Etrans = 0.5￼ = 0.5 x 0.2 x 0.82 = 0.064 J So the difference in energy would be 0.078J - 0.064J = 0.014J so the velocity of m1 after the collision v = √(0.014J) / (0.5 x 0.08kg) = 0.6ms-1 ---------------- My answer just seems a long winded way of doing it but for 9 marks I can't see where I've gone wrong if I have. It feels wrong because v1 of m1 is just the difference between the initial velocity of the m1 and the final velocity of m2... Any pointers would be greatly appreciated.