Two perfectly elastic particles P and Q of equal mass travelling along the line joining them with velocities \[15{\text{ }}m/sec\] and \[10{\text{ }}m/sec\]. After collision, their velocities respectively (in m/sec) will be.
A. $0,25$
B. $5,20$
C. $10,15$
D. $20,5$
Answer
299.7k+ views
Hint:In this question, we are given that there are two perfectly elastic particles P and Q with the initial velocities \[15{\text{ }}m/sec\] and \[10{\text{ }}m/sec\]. Also, the masses of both the particles are the same. We have to calculate their velocities after collision. To calculate the velocities, we’ll use the formula of velocities after collision (elastic bodies).
Formula used:
When two bodies collides (elastically) then the velocities are;
${v_1} = \dfrac{{\left( {{m_1} - {m_2}} \right){u_1} + 2{m_2}{u_2}}}{{{m_1} + {m_2}}}$, ${v_2} = \dfrac{{\left( {{m_2} - {m_1}} \right){u_2} + 2{m_1}{u_1}}}{{{m_1} + {m_2}}}$
Here, ${m_1},{m_2}$ and ${u_1},{u_2}$ are the masses and initial velocities of both the bodies respectively.
Complete step by step solution:
Given that,
Velocities of particles P and Q are \[15{\text{ }}m/sec\] and \[10{\text{ }}m/sec\] respectively.
Initial velocity of P, ${u_1} = 15{\text{ }}m/sec$
Initial velocity of Q, ${u_2} = 10{\text{ }}m/sec$
The Masses of the particles P and Q are the same.
Therefore, let ${m_1} = {m_2} = m$
After collision their velocities will be,
Using the formula,
${v_1} = \dfrac{{\left( {{m_1} - {m_2}} \right){u_1} + 2{m_2}{u_2}}}{{{m_1} + {m_2}}} \\ $
$\Rightarrow {v_2} = \dfrac{{\left( {{m_2} - {m_1}} \right){u_2} + 2{m_1}{u_1}}}{{{m_1} + {m_2}}} \\ $
Put the value of ${u_1} = 15{\text{ }}m/sec$, ${u_2} = 10{\text{ }}m/sec$ and ${m_1} = {m_2} = m$
It will be, velocity of P, ${v_1} = \dfrac{{\left( {m - m} \right)15 + 2m\left( {10} \right)}}{{m + m}}$
${v_1} = \dfrac{{20m}}{{2m}}$
$\Rightarrow {v_1} = 10{\text{ }}m/sec$
And Velocity of Q,
${v_2} = \dfrac{{\left( {m - m} \right)10 + 2m\left( {15} \right)}}{{m + m}} \\ $
$\Rightarrow {v_2} = \dfrac{{30m}}{{2m}} \\ $
$\therefore {v_2} = 15{\text{ }}m/sec$
Hence, option C is the correct answer.
Note: To solve such problems, one should always remember the concept of collision of elastic bodies that if the mass is the same and the bodies are perfectly elastic then after collision their velocities get interchanged. Consider the two particles or bodies of equal mass moving in opposite directions at the same velocity. In this case, both bodies come to rest after the collision, implying that no velocity exchange occurs.
Formula used:
When two bodies collides (elastically) then the velocities are;
${v_1} = \dfrac{{\left( {{m_1} - {m_2}} \right){u_1} + 2{m_2}{u_2}}}{{{m_1} + {m_2}}}$, ${v_2} = \dfrac{{\left( {{m_2} - {m_1}} \right){u_2} + 2{m_1}{u_1}}}{{{m_1} + {m_2}}}$
Here, ${m_1},{m_2}$ and ${u_1},{u_2}$ are the masses and initial velocities of both the bodies respectively.
Complete step by step solution:
Given that,
Velocities of particles P and Q are \[15{\text{ }}m/sec\] and \[10{\text{ }}m/sec\] respectively.
Initial velocity of P, ${u_1} = 15{\text{ }}m/sec$
Initial velocity of Q, ${u_2} = 10{\text{ }}m/sec$
The Masses of the particles P and Q are the same.
Therefore, let ${m_1} = {m_2} = m$
After collision their velocities will be,
Using the formula,
${v_1} = \dfrac{{\left( {{m_1} - {m_2}} \right){u_1} + 2{m_2}{u_2}}}{{{m_1} + {m_2}}} \\ $
$\Rightarrow {v_2} = \dfrac{{\left( {{m_2} - {m_1}} \right){u_2} + 2{m_1}{u_1}}}{{{m_1} + {m_2}}} \\ $
Put the value of ${u_1} = 15{\text{ }}m/sec$, ${u_2} = 10{\text{ }}m/sec$ and ${m_1} = {m_2} = m$
It will be, velocity of P, ${v_1} = \dfrac{{\left( {m - m} \right)15 + 2m\left( {10} \right)}}{{m + m}}$
${v_1} = \dfrac{{20m}}{{2m}}$
$\Rightarrow {v_1} = 10{\text{ }}m/sec$
And Velocity of Q,
${v_2} = \dfrac{{\left( {m - m} \right)10 + 2m\left( {15} \right)}}{{m + m}} \\ $
$\Rightarrow {v_2} = \dfrac{{30m}}{{2m}} \\ $
$\therefore {v_2} = 15{\text{ }}m/sec$
Hence, option C is the correct answer.
Note: To solve such problems, one should always remember the concept of collision of elastic bodies that if the mass is the same and the bodies are perfectly elastic then after collision their velocities get interchanged. Consider the two particles or bodies of equal mass moving in opposite directions at the same velocity. In this case, both bodies come to rest after the collision, implying that no velocity exchange occurs.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

