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one dimension collisions

A ball of mass 1.5kg is thrown upward. it leaves the thrower's hand with a velocity of 10 m/s. Assuming the upward direction is positive:
a. how long does it take for the ball to return to the throwers hand

b.what is the ball's final velocity just before it reaches the hand?

c. what is the change in momentum of the ball?

d. what is the impulse calculated from the change in momentum?

e. what is the avg force acting on the ball?

f. After the ball hits the throwers hand, it comes to rest in 0.25s. What is the net impulse of the ball and what is the avg force exerted by the hand on the ball?

Solution Preview

m=mass of ball= 1.5 kg
initial velocity = u= 10 m/s

a. how long does it take for the ball to return to the throwers hand

x= ut + 1/2 at^2

u= 10 m/s
a=acceleration due to gravity= -9.81 m/s^2
(it is negative because it is acting downwards, upwards is positive)

x= 0 (as it returns to thrower's hand)

Therefore, x= ut + 1/2 at^2 = 0

10t - 1/2 x 9.81 x t^2 = 0
or t= 2.04 s

Answer: 2.04 s

b.what is the ball's final ...

Solution Summary

Answers to questions on one dimension collisions.

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