Question

If the ball has lost half the magnitude of its impact momentum immediately after it recoils, to what height does the ball reach after its first rebound? 1. h 2. h 3 3. 0 4. h 2 5. h 4

Answer 1

See below.

Explanation:

Let the ball be dropped from a height H.

therefore,
`u=0`
`a=-g`
`S=-H`

where `u` is initial velocity, `a` is acceleration due to gravity, `S` equals distance traveled.

So,
`v^2 = u^2 + 2aS`
`v^2 = 2(-g)(-H)`
`v = sqrt(2gH)`

(You could have also derived this using energy conservation)

Let the mass of the ball be `m`.

initial momentum = `msqrt(2gH)`

final momentum = `1/2`initial momentum
= `(msqrt(2gH))/2`

We consider the mass of the ball to remain same

So,

`v_o =(sqrt(2gH))/2`

where `v_o` equals the velocity of the ball immediately after the impact.

Since we want to find the height,

`u_o = (sqrt(2gH))/2`
`a = -g`
`v_o = 0`

where `u_o` is the initial velocity after the impact, `a` is acceleration due to gravity, `v_o` is final velocity which is zero since the ball is moving upwards.

`v_o^2 = u_o^2 + 2aS`
`0 = (gH)/2 + 2(-g)(S)`
`S = H/4`

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