
A body of mass $m$ is projected at an angle of \[45^\circ \] with the horizontal. If air resistance is negligible, then what will be the total change in momentum when it strikes the ground?
Answer
150.3k+ views
Hint: In this solution, we will first calculate the component of the projectile body perpendicular to the surface of the Earth when it strikes the ground. The ball after striking the ground will maintain the magnitude of its velocity but reverse its direction.
Complete step by step answer:
We’ve been told a projectile is launched at an angle of \[45^\circ \] with the horizontal. When the projectile is about to complete its motion and collide back with the ground, all its energy will be in the form of kinetic energy when it strikes the ground. This implies that the velocity of the projectile will have the same value when it was launched.
However, the perpendicular component of the velocity will be downwards since the ball will be approaching the ground. If the projectile of the velocity is denoted by $v$, its component perpendicular to the ground will be $v\sin \theta $.
Now when the ball strikes the ground it will bounce back up. But this time, it will have a perpendicular velocity in the upwards direction that is opposite to its initial direction when it strikes the ground. So, we can calculate the net change in momentum as
$\Delta P = mv\sin \theta - ( - mv\sin \theta )$
$ \Rightarrow \Delta P = 2mv\sin \theta $
Since \[\theta = 45^\circ \],
$\Delta P = 2mv\dfrac{1}{{\sqrt 2 }}$
Which can be simplified to
$\Delta P = \dfrac{{mv}}{{\sqrt 2 }}$
Note: Here we have assumed that the ball will not lose its energy when it collides with the ground. In reality, there will always be a loss of energy when the ball collides with the ground and hence its velocity will be lower than the initial velocity it had when striking the ground.
Complete step by step answer:
We’ve been told a projectile is launched at an angle of \[45^\circ \] with the horizontal. When the projectile is about to complete its motion and collide back with the ground, all its energy will be in the form of kinetic energy when it strikes the ground. This implies that the velocity of the projectile will have the same value when it was launched.
However, the perpendicular component of the velocity will be downwards since the ball will be approaching the ground. If the projectile of the velocity is denoted by $v$, its component perpendicular to the ground will be $v\sin \theta $.
Now when the ball strikes the ground it will bounce back up. But this time, it will have a perpendicular velocity in the upwards direction that is opposite to its initial direction when it strikes the ground. So, we can calculate the net change in momentum as
$\Delta P = mv\sin \theta - ( - mv\sin \theta )$
$ \Rightarrow \Delta P = 2mv\sin \theta $
Since \[\theta = 45^\circ \],
$\Delta P = 2mv\dfrac{1}{{\sqrt 2 }}$
Which can be simplified to
$\Delta P = \dfrac{{mv}}{{\sqrt 2 }}$
Note: Here we have assumed that the ball will not lose its energy when it collides with the ground. In reality, there will always be a loss of energy when the ball collides with the ground and hence its velocity will be lower than the initial velocity it had when striking the ground.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Charging and Discharging of Capacitor

Ideal and Non-Ideal Solutions Raoult's Law - JEE

The time period of an artificial satellite in a circular class 11 physics JEE_Main

JEE Main Chemistry Question Paper with Answer Keys and Solutions

A pressure of 100 kPa causes a decrease in volume of class 11 physics JEE_Main

Other Pages
List of Fastest Century In IPL - Cricket League and FAQs

JEE Main Response Sheet 2025 Released – Download Links, and Check Latest Updates

JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

NEET Physics Syllabus 2025 FREE PDF: Important Topics and Weightage

List of 10 Fastest Centuries in ODIs - Players and Countries

NCERT Solutions for Class 9 Maths Updated for 2024-25 Exam | Free PDF
