
Two solid rubber balls, A and B having masses $200g$ and $400g$ respectively are moving in opposite directions with velocity of A equal to $0.3\,m{s^{ - 1}}$. After collision the two balls come to rest, then the velocity of B is-
A. $0.15\,m{s^{ - 1}}$
B. $1.5\,m{s^{ - 1}}$
C. $ - 0.15\,m{s^{ - 1}}$
D. None of the above
Answer
232.8k+ views
Hint:In order to solve this question, we will apply the law of conservation of momentum and then using this principle and equations we will determine the velocity of ball B before the collision.
Formula used:
The principle of conservation of linear momentum says that Initial momentum of a system is always equal to final momentum of a system. ${P_i} = {P_f}$ where $P = mv$ denotes the momentum of a body defined as the product of mass of the body and the velocity of the body.
Complete step by step solution:
According to the question, we have given that mass of body A and B are ${m_A} = 200g = 0.2kg$ and ${m_B} = 400g = 0.4kg$ and initial velocity of body A was ${u_A} = 0.3m{s^{ - 1}}$ and let initial velocity of B was ${u_B}$ so total initial momentum of the system was;
${P_i} = {m_A}{u_A} + {m_B}{u_B} \\
\Rightarrow {P_i} = 0.2(0.3) + 0.4({u_B}) \to (i) \\ $
After collision, both bodies came to rest which means final velocities of both bodies is zero so final momentum will be zero ${P_f} = 0 \to (ii)$ So, using principle of conservation of linear momentum we have,
${P_i} = {P_f}$
So using equations (i) and (ii) we get,
$0.06 + 0.4\,{u_B} = 0 \\
\Rightarrow {u_B} = \dfrac{{ - 0.06}}{{0.4}} \\
\therefore {u_B} = - 0.15\,m{s^{ - 1}} \\ $
Hence, the correct answer is option C.
Note: Other than the law of conservation of linear momentum, the other two most significant conservation laws are the law of conservation of energy, which states that the system's energy is preserved, and the law of conservation of angular momentum in rotational dynamics.
Formula used:
The principle of conservation of linear momentum says that Initial momentum of a system is always equal to final momentum of a system. ${P_i} = {P_f}$ where $P = mv$ denotes the momentum of a body defined as the product of mass of the body and the velocity of the body.
Complete step by step solution:
According to the question, we have given that mass of body A and B are ${m_A} = 200g = 0.2kg$ and ${m_B} = 400g = 0.4kg$ and initial velocity of body A was ${u_A} = 0.3m{s^{ - 1}}$ and let initial velocity of B was ${u_B}$ so total initial momentum of the system was;
${P_i} = {m_A}{u_A} + {m_B}{u_B} \\
\Rightarrow {P_i} = 0.2(0.3) + 0.4({u_B}) \to (i) \\ $
After collision, both bodies came to rest which means final velocities of both bodies is zero so final momentum will be zero ${P_f} = 0 \to (ii)$ So, using principle of conservation of linear momentum we have,
${P_i} = {P_f}$
So using equations (i) and (ii) we get,
$0.06 + 0.4\,{u_B} = 0 \\
\Rightarrow {u_B} = \dfrac{{ - 0.06}}{{0.4}} \\
\therefore {u_B} = - 0.15\,m{s^{ - 1}} \\ $
Hence, the correct answer is option C.
Note: Other than the law of conservation of linear momentum, the other two most significant conservation laws are the law of conservation of energy, which states that the system's energy is preserved, and the law of conservation of angular momentum in rotational dynamics.
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

