
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
232.8k+ 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.
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

