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

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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

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Answer 1 · 2017-12-11T19:28:17

See below.

Explanation:

Let the ball be dropped from a height H.

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

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

So,
$v^{2} = u^{2} + 2 a S$ $v^2 = u^2 + 2aS$
$v^{2} = 2 (- g) (- H)$ $v^2 = 2(-g)(-H)$
$v = \sqrt{2 g H}$ $v = sqrt(2gH)$

(You could have also derived this using energy conservation)

Let the mass of the ball be $m$ $m$ .

initial momentum = $m \sqrt{2 g H}$ $msqrt(2gH)$

final momentum = $\frac{1}{2}$ $1/2$ initial momentum
= $\frac{m \sqrt{2 g H}}{2}$ $(msqrt(2gH))/2$

We consider the mass of the ball to remain same

So,

$v_{o} = \frac{\sqrt{2 g H}}{2}$ $v_o =(sqrt(2gH))/2$

where $v_{o}$ $v_o$ equals the velocity of the ball immediately after the impact.

Since we want to find the height,

$u_{o} = \frac{\sqrt{2 g H}}{2}$ $u_o = (sqrt(2gH))/2$
$a = - g$ $a = -g$
$v_{o} = 0$ $v_o = 0$

where $u_{o}$ $u_o$ is the initial velocity after the impact, $a$ $a$ is acceleration due to gravity, $v_{o}$ $v_o$ is final velocity which is zero since the ball is moving upwards.

$v_{o}^{2} = u_{o}^{2} + 2 a S$ $v_o^2 = u_o^2 + 2aS$
$0 = \frac{g H}{2} + 2 (- g) (S)$ $0 = (gH)/2 + 2(-g)(S)$
$S = \frac{H}{4}$ $S = H/4$

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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

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