1.) Your friend is catching a falling basketball after it has passed through the

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.)A 70-kg astronaut pushes against the inside back wall of a 2000-kg spaceship and moves toward the front. Her speed increases from 0 to 2.4 m/s. Earth is the object of reference.

If her push lasts 0.30 s, what is the average force that the astronaut exerts on the wall of the spaceship?

If the spaceship was initially at rest, with what speed does it recoil?

3.) 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.

Explanation / Answer

1)

Velocity of ball when it touches the hand = sqrt(2gh) = 4.8m/s

acc on the ball by hand = 4.8/t = 4.8/0.1 = 48m/s2

force exerted on the hand = 0.6*a = 28.8 N

2)

avg acc= 2.4/0.3 = 8m/s2

avg force = 70*8 = 560N

recoil speed = 70*2.4/2000 = 0.084 m/s

3)

Ma = 0.248 kg Ua =? Vb = -0.125m/s

Mb = 0.372 kg Ub = 0m/s Vb = 0.653m/s

Ma*Ua + Mb*Ub = Ma*Va +Mb*Vb

0.248*Ua + 0 = 0.248*(-0.125) + 0.372*0.653

Ua = 0.854 m/s towards right

Chage in kinetic energy = 0.5Ma*Ua2 + 0.5Mb*Ub2 - (0.5Ma*Va2 + 0.5Mb*Vb2)

= 0.0183 J

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