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Part C) only A light inextensible string passes over a smooth fixed pulley. One end of...

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roshan-rox | (Level 1) Valedictorian

Posted September 9, 2013 at 11:13 AM via web

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Part C) only

A light inextensible string passes over a smooth fixed pulley. One end of the string carries a bucket of mass M and the other end carries a counterpoise of equal mass. A small ball ball of mass m is dropped vertically so as to strike the horizontal bottom of the bucket with velocity u. If e is the coefficient of restitution,

A)

show that the bucket begins to move with velocity (m(1+e)u)/(2M+m)

B)

find the impulse in the string.

C)

Find , also, the time between the first and second impacts of the ball and the bucket.

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted October 14, 2013 at 3:11 PM (Answer #1)

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A)

The figure is below. Initially the bucket and the counterweight are stationary. After the small ball bounces back, it will have the speed -`e*u` relative to the bucked which will move with the speed `v` . Thus the total speed of the ball after first bounce will be `v-eu`.

In all collisions the linear momentum is the same before and after collision. Initially only the ball has a momentum. Finally both the ball (moving with speed `v-eu`)  and the system of bucket plus counterweight (which moves with speed `v`) have momentum.

`m*u = m*(v-eu) +2M*v`

`m*u(1+e) =mv +2Mv`

`v = [m*(1+e)*u]/(m+2M)`

B)

The force in the string is (as momentum theorem says):

`F = (Delta(P))/(Delta(t)) =(M*v)/(Delta(t))`

Thus the momentum in string is just

`P = F*Delta(t) =M*v = [Mm*(1+e)]/(m+ 2M) *u`

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted October 14, 2013 at 3:21 PM (Answer #2)

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Part C) is an entire new question and that is why there are two answers here to the problem.

Taking the positive axis downwards, the distance that the small ball travels is

`s = (v-eu)*t +(g*t^2)/2`

Also this distance for the bucket that has a speed `v` after collision is

`s =v*t`

By equating the above two expressions one gets

`(v-eu)*t +(g*t^2)/2 =v*t`

`(g*t^2)/2 -eu*t =0`

`t*(g*t/2 -e*u) =0`

This equation has two solutions

`t=0` initial time when the ball bounces first from the bucket

`t = (2eu)/g` time when the ball bounces second from the bucket.

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