Question

1.) Your friend is catching a falling basketball after it has passed through the basket. Her hands move straight down while catching the ball. Take that it takes about 0.10 s for the player to lower her hands to stop the ball. Assume the mass of the ball to be 0.60 kg, and that it has fallen a vertical distance of 1.2 m before reaching the player's hand. Determine the average force that her hands exert on the ball while catching it.

2.) On a frictionless horizontal air table, puck A (with mass 0.248kg ) is moving toward puck B (with mass 0.372kg ), which is initially at rest. After the collision, puck A has velocity 0.125m/s to the left, and puck B has velocity 0.653m/s to the right.

What was the speed vAi of puck A before the collision?

Calculate ?K, the change in the total kinetic energy of the system that occurs during the collision.

Answer 1

i)        acceleration due to gravity, g = 9.8 m/s²

Velocity of the ball when it touches the hands, V_i = sqrt(2*g*h) = sqrt(2*9.8*1.2) = 4.85 m/s

Final velocity of the ball, V_f = 0

Time taken by the ball to stop, t= 0.1s

Average decceleration of the ball while being stopped, a = (V_i - V_f)/t = 4.85/0.1 = 48.5 m/s²

Force acting on the ball, F = m*a = 0.6*48.5 = 29.1 N

ii)     m_A = 0.248 kg, m_B = 0.372 kg, V_Ai = to be found (+ve to the right), V_Bi = 0, V_Af = - 0.125 m/s to left, V_Bf = 0.653 m/s to right

By Conservation of Momentum

m_A*V_Ai + m_B*V_Bi = m_A*V_Af + m_B*V_Bf

V_Ai= (m_A*V_Af + m_B*V_Bf)/m_A = (- 0.248*0.125 +0.372*0.653)/0.248 = 0.8545 m/s

Initial Total K.E. = 0.5*m_A*V_Ai² + 0.5*m_B*V_Bi² = 0.5*0.248*0.8545² = 0.09054J

Final Total K.E. = 0.5*m_A*V_Af² + 0.5*m_B*V_Bf² = 0.5*0.248*0.125 + 0.5*0.372*0.653 = 0.08125 J

Change in Total K.E. = 0.09054 - 0.08125 = 0.00929 J