Physics: force and motion

code PLUS

2012-06-07 12:59 am

Take g=10 m s-2.

A ball of mass 50 g, which is initially at rest, is dropped onto the ground from a height of 1 m. If it rises back to a height of 0.8 m after the rebound, find the change in its momentum during the rebound.

A. 0.236 kg m s-1
B. 0.250 kg m s-1
C. 0.424 kg m s-1
D. 0.472kg m s-1

My steps are as followed:

(0.5mv^2)+(mgh)=(0.5mv^2)+(mgh)
i.e. 0+(0.05)(10)(1) =(0.5)(0.05)(v^2)+(0.05)(10)(0.8)
v=2 m s-1

mv-mu=(0.05)(2)-(0.05)(0)
=0.01 kg m s-1

Please point out my fault and give me the correct steps.

回答 (1)

天同

2012-06-07 1:34 am

✔ 最佳答案

The speed of the ball before striking the ground
= square-root[2g x 1] m/s = 4.472 m/s
where g is the acceleration due to gravity, taken to be 10 m/s2

The speed of the ball immediately after rebound
= square-root[2g x 0.8] m/s = 4 m/s

Hence, change in momentum
= final momentum - initial momentum
= 0.05 x 4 - (0.05 x (-4.472)) kg/m/s (upward direction is taken as +ve)
= 0.4236 kg.m/s
Option C is the answer.

Your equations are all wrong.

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