Question

On a frictionless. horizontal air table, puck A (with mass 0.250 kg) is moving toward puck B (with mass 0.360 kg), that is initially at rest. After the collision, puck A has a velocity of 0.121 m/s to the left, and puck B has velocity 0.660 m/s to the right. Part A What was the speed of puck A before the collision?

Answer 1

Answer for part b will be 5.749×10^-3 J.

-ve means loss in kinetic energy.

After Collision, final momentum, P_f = m_B v_B - m_A v_A [moving in opposite direction] = [0.36 times 0.66 - 0.25 times 0.121] P_f = 0.20735 kgm/s. (towords right) As there is no external force, moment conservation, P_i = P_f P_i (before = 0.20735 kgm/s. collision) Hence, P_i = m_A mu_A Implies u_A = 0.8294 m.s (rightwords) \r\n Hence, P_i = m_A mu_A Implies u_A = 0.8294 m/s (rightwords) change in K.E Implies K.E_f - K.E_i Implies 1/2 m_A u^2_A + 1/2 m_B u^2_B - 1/2 m_A u^2_A = 1/2 times 0.25 times 0.121^2 + 1/2 times 0.36 times 0.66&2 - 1/2 times 0.25 times 0.8294^2 = -0.00574992 J (-ve means loss in K.E)