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

1 Answer
Dec 11, 2017

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,
v2=u2+2aS
v2=2(g)(H)
v=2gH

(You could have also derived this using energy conservation)

Let the mass of the ball be m.

initial momentum = m2gH

final momentum = 12initial momentum
= m2gH2

We consider the mass of the ball to remain same

So,

vo=2gH2

where vo equals the velocity of the ball immediately after the impact.

Since we want to find the height,

uo=2gH2
a=g
vo=0

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

v2o=u2o+2aS
0=gH2+2(g)(S)
S=H4

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