Question

A tennis ball of mass 0.063kg is held just above a basketball of mass 0.530kg. With their centers vertically aligned, both balls are released from rest at the same time and fall through a distance of 1.5m. 1) How fast are the balls moving right before impact? 2) How fast are they moving after impact? Assume an elastic collision 3) How high will be tennis ball travel after collision? ) What is the impulse delivered to the tennis ball? 5) If the collision takes places over a time interval of 0.16s, assuming a constant force, what is the acceleration of the tennis ball during the collision?

Answer 1

"Conservation of Energy (g = 9.81 m/s^2) mgh = 1/2 mv^2 V = squareroot 2gh V = 5.40 m/s 1 # Basket Ball 2 # Tennis Ball let velocities after impact are V_1 & V_2 Basket Ball rebounds with same velocity (-5.40 m/s) while Tennis Ball is travelling at (+5.40 m/s) Let these velocities be u_1 & u_2 1^st conservation of momentum m_1u_1 + m_2u_2 = m_1V_1 + m_2V_2 0.530 times (-5.40) + 0.063 times 5.40 = 0.530 V_1 + 0.063 V_2 -2.52 = 0.530V_1 + 0.063 V_2 (1) 2^nd conservation of Energy 1/2 m_1V_1^2 + 1/2 m_2V_2^2 = 1/2 m_1u_1^2 + 1/2 m_2u_2^2 0.530 V_1^2 + 0.063 V_2^2 = 17.29 (2) Putting V_2 = -(0.530V_1 + 2.52)/0.063 from (1) in (2) we get 4.99 V_1^2 + 42.4 V_1 + 83.51 = 0 V_1 = -3.1 m/s V_2 = -13.92 m/s means Basket Ball Travelling up at 3.1 m/s