
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
220.5k+ 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
Two discs which are rotating about their respective class 11 physics JEE_Main

A ladder rests against a frictionless vertical wall class 11 physics JEE_Main

Two simple pendulums of lengths 1 m and 16 m respectively class 11 physics JEE_Main

The slopes of isothermal and adiabatic curves are related class 11 physics JEE_Main

A trolly falling freely on an inclined plane as shown class 11 physics JEE_Main

The masses M1 and M2M2 M1 are released from rest Using class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

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

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

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

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

