
Particle A makes a head on elastic collision with another stationary particle B. They fly apart in opposite directions with equal speeds. The mass ratio will be-
A. \[\dfrac{1}{3}\]
B. \[\dfrac{1}{2}\]
C. \[\dfrac{1}{4}\]
D. \[\dfrac{2}{3}\]
Answer
144.6k+ views
Hint:According to the law of conservation of momentum, in an isolated system the total momentum of bodies acting upon each other does not change or remains constant unless an external force is applied to it. So, we can say that momentum can neither be created nor destroyed. Hence momentum before the collision is equal to the momentum after the collision. By using this we can find the relation between the bodies.
Formula used:
According to the conservation of momentum,
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Where \[{u_1}\] and \[u_2\] are initial velocities and \[v_1\] and \[v_2\] are initial velocities.
Complete step by step solution:
For an elastic collision,
\[e = - \dfrac{{{v_1} - {v_2}}}{{{u_1} - {u_2}}} = 1\]
Let before collision conditions, the velocity of A be \[{u_1}\] and the velocity of B be \[{u_2} = 0\]. After collision conditions, the speed of A be \[{v_1}\] and the speed of B be\[{v_2}\].
According to the conservation of momentum,
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Using \[{u_2} = 0\]
\[{m_1}{u_1} = {m_1}{v_1} - {m_2}{v_1}\] ……(1)
As we know that \[e = - \dfrac{{{v_1} - {v_2}}}{{{u_1} - {u_2}}} = 1\]
\[{v_2} - ( - {v_1}) = {u_1} - {u_2}\]
\[\Rightarrow 2{v_1} = {u_1}\] ……(2)
Putting equation 2 in 1, we get
\[2{m_1}{v_1} = {m_2}{v_1} - {m_1}{v_1}\]
\[\Rightarrow 3{m_1}{v_1} = {m_2}{v_1}\]
\[\therefore \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{3}\]
Therefore, the mass ratio will be \[\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{3}\].
Hence option A is the correct answer.
Note: When two bodies come in direct contact, then a collision must be occurring. Collision is the event in which the bodies exert forces on one another in about a relatively short time. An elastic collision is defined as one where there is no net loss of kinetic energy in the system due to the collision. A perfectly elastic collision is when two bodies collide but there is no loss in the total kinetic energy.
Formula used:
According to the conservation of momentum,
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Where \[{u_1}\] and \[u_2\] are initial velocities and \[v_1\] and \[v_2\] are initial velocities.
Complete step by step solution:
For an elastic collision,
\[e = - \dfrac{{{v_1} - {v_2}}}{{{u_1} - {u_2}}} = 1\]
Let before collision conditions, the velocity of A be \[{u_1}\] and the velocity of B be \[{u_2} = 0\]. After collision conditions, the speed of A be \[{v_1}\] and the speed of B be\[{v_2}\].
According to the conservation of momentum,
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Using \[{u_2} = 0\]
\[{m_1}{u_1} = {m_1}{v_1} - {m_2}{v_1}\] ……(1)
As we know that \[e = - \dfrac{{{v_1} - {v_2}}}{{{u_1} - {u_2}}} = 1\]
\[{v_2} - ( - {v_1}) = {u_1} - {u_2}\]
\[\Rightarrow 2{v_1} = {u_1}\] ……(2)
Putting equation 2 in 1, we get
\[2{m_1}{v_1} = {m_2}{v_1} - {m_1}{v_1}\]
\[\Rightarrow 3{m_1}{v_1} = {m_2}{v_1}\]
\[\therefore \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{3}\]
Therefore, the mass ratio will be \[\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{1}{3}\].
Hence option A is the correct answer.
Note: When two bodies come in direct contact, then a collision must be occurring. Collision is the event in which the bodies exert forces on one another in about a relatively short time. An elastic collision is defined as one where there is no net loss of kinetic energy in the system due to the collision. A perfectly elastic collision is when two bodies collide but there is no loss in the total kinetic energy.
Recently Updated Pages
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

Charle's Law Formula - Definition, Derivation and Solved Examples

Central Angle of a Circle Formula - Definition, Theorem and FAQs

Average Force Formula - Magnitude, Solved Examples and FAQs

Boyles Law Formula - Boyles Law Equation | Examples & Definitions

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

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

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
