
A ball of mass $m$ falls vertically to the ground from a height ${h_1}$ and rebound to a height ${h_2}$. The change in momentum of the ball on striking ground is:
A. $mg\left( {{h_1} - {h_2}} \right) \\ $
B. $m\left( {\sqrt {2g{h_2}} + \sqrt {2g{h_1}} } \right) \\ $
C. $m\sqrt {2g\left( {{h_1} + {h_2}} \right)} \\ $
D. $m\sqrt {2g} \left( {{h_1} + {h_2}} \right)$
Answer
162.6k+ views
Hint:In this question, we are given the mass of the ball. Ball is falling vertically to the ground and rebound (height is given). Apply the formula of velocity to calculate initial and final velocity. Then, apply the formula of change in momentum.
Formula used:
Change in momentum –
\[\Delta P = mv - mu\]
Here, $v = $ final velocity
$u = $ initial velocity
Velocity of an object falling from the height $h$ –
$v = \sqrt {2gh} $
Complete step by step solution:
Given that, the mass of the ball is $m$. Now, the ball is falling vertically from the height of ${h_1}$ to the ground and then rebounds at the height ${h_2}$. Therefore, the initial height is ${h_1}$ and the final height is ${h_2}$. Applying the formula of velocity i.e.,
$v = \sqrt {2gh} $
Thus, the initial velocity $u = \sqrt {2g{h_1}} $
Initial velocity will be negative because the ball is falling in the downward direction.
It implies that,
$u = - \sqrt {2g{h_1}} $
And the final velocity,
$v = \sqrt {2g{h_2}} $
As we know, impulse is equal to the change in momentum and momentum is the product of mass and the velocity of the body. Also, change in momentum is the difference between final momentum and the initial momentum.
$\text{Change in the momentum} = mv - mu \\ $
$\Rightarrow \text{Change in the momentum} = m\sqrt {2g{h_2}} - \left( { - m\sqrt {2g{h_1}} } \right) \\ $
$\therefore \text{Change in the momentum} = m\left( {\sqrt {2g{h_2}} + \sqrt {2g{h_1}} } \right)$
Hence, option B is the correct answer .
Note: There is a third approach to kinetics difficulties in dynamics, which is provided by the ideas of impulse and momentum. The Impulse-momentum method is based on the notion that the force applied to a body over a period of time is equivalent to the change in momentum applied to that body.
Formula used:
Change in momentum –
\[\Delta P = mv - mu\]
Here, $v = $ final velocity
$u = $ initial velocity
Velocity of an object falling from the height $h$ –
$v = \sqrt {2gh} $
Complete step by step solution:
Given that, the mass of the ball is $m$. Now, the ball is falling vertically from the height of ${h_1}$ to the ground and then rebounds at the height ${h_2}$. Therefore, the initial height is ${h_1}$ and the final height is ${h_2}$. Applying the formula of velocity i.e.,
$v = \sqrt {2gh} $
Thus, the initial velocity $u = \sqrt {2g{h_1}} $
Initial velocity will be negative because the ball is falling in the downward direction.
It implies that,
$u = - \sqrt {2g{h_1}} $
And the final velocity,
$v = \sqrt {2g{h_2}} $
As we know, impulse is equal to the change in momentum and momentum is the product of mass and the velocity of the body. Also, change in momentum is the difference between final momentum and the initial momentum.
$\text{Change in the momentum} = mv - mu \\ $
$\Rightarrow \text{Change in the momentum} = m\sqrt {2g{h_2}} - \left( { - m\sqrt {2g{h_1}} } \right) \\ $
$\therefore \text{Change in the momentum} = m\left( {\sqrt {2g{h_2}} + \sqrt {2g{h_1}} } \right)$
Hence, option B is the correct answer .
Note: There is a third approach to kinetics difficulties in dynamics, which is provided by the ideas of impulse and momentum. The Impulse-momentum method is based on the notion that the force applied to a body over a period of time is equivalent to the change in momentum applied to that body.
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