
A hockey ball of mass 200 g travelling at $10m{s^{ - 1}}$ is struck by a hockey stick so as to return it along its original path with a velocity of$5m{s^{ - 1}}$. Calculate the change of momentum which occurred in the motion of the hockey ball by the force applied by the hockey stick:
A) -3 Ns.
B) 3 Ns.
C) 6 Ns.
D) -6 Ns.
Answer
241.2k+ views
Hint: Momentum is defined as the product of mass and velocity. The rate of change of momentum with respect to time is known as force. The force is defined as the product of mass and acceleration of the body and unit of force is Newtons.
Formula Used: The formula of momentum is given by,
$ \Rightarrow p = m \cdot v$
Where p is the momentum the mass is m and the velocity is v.
Complete step by step answer:
It is given in the problem that a hockey ball of mass 200 g travelling at $10m{s^{ - 1}}$ is struck by a hockey stick so as to return it along its original path with a velocity of $5m{s^{ - 1}}$ and we need to calculate the change in the momentum of the hockey ball.
The formula of momentum is given by,
$ \Rightarrow p = m \cdot v$
Where p is the momentum the mass is m and the velocity is v.
The mass of the hockey ball is 200 g and the initial velocity of the ball is $10m{s^{ - 1}}$ and the final velocity of the ball is $5m{s^{ - 1}}$ in the opposite direction of the initial direction. The change in the momentum is given by,
$\Delta p = m\left( {v - u} \right)$
Where $\Delta p$ is change in momentum mass is m initial and final velocity of the hockey ball is u and v respectively.
The change in momentum of hockey ball is equal to,
$ \Rightarrow \Delta p = m\left( {v - u} \right)$
$ \Rightarrow \Delta p = \left( {\dfrac{{200}}{{1000}}} \right) \cdot \left[ {5 - \left( { - 10} \right)} \right]$
\[ \Rightarrow \Delta p = \left( {0 \cdot 2} \right) \cdot \left( {15} \right)\]
\[ \Rightarrow \Delta p = 3Ns\]
The change in the momentum is\[\Delta p = 3Ns\].
The correct answer for this problem is option B.
Note: Momentum is a vector quantity and therefore if there is change in line of motion of the body then we calculate the momentum along the same directions. The velocity of the hockey ball is negative while returning towards its original path.
Formula Used: The formula of momentum is given by,
$ \Rightarrow p = m \cdot v$
Where p is the momentum the mass is m and the velocity is v.
Complete step by step answer:
It is given in the problem that a hockey ball of mass 200 g travelling at $10m{s^{ - 1}}$ is struck by a hockey stick so as to return it along its original path with a velocity of $5m{s^{ - 1}}$ and we need to calculate the change in the momentum of the hockey ball.
The formula of momentum is given by,
$ \Rightarrow p = m \cdot v$
Where p is the momentum the mass is m and the velocity is v.
The mass of the hockey ball is 200 g and the initial velocity of the ball is $10m{s^{ - 1}}$ and the final velocity of the ball is $5m{s^{ - 1}}$ in the opposite direction of the initial direction. The change in the momentum is given by,
$\Delta p = m\left( {v - u} \right)$
Where $\Delta p$ is change in momentum mass is m initial and final velocity of the hockey ball is u and v respectively.
The change in momentum of hockey ball is equal to,
$ \Rightarrow \Delta p = m\left( {v - u} \right)$
$ \Rightarrow \Delta p = \left( {\dfrac{{200}}{{1000}}} \right) \cdot \left[ {5 - \left( { - 10} \right)} \right]$
\[ \Rightarrow \Delta p = \left( {0 \cdot 2} \right) \cdot \left( {15} \right)\]
\[ \Rightarrow \Delta p = 3Ns\]
The change in the momentum is\[\Delta p = 3Ns\].
The correct answer for this problem is option B.
Note: Momentum is a vector quantity and therefore if there is change in line of motion of the body then we calculate the momentum along the same directions. The velocity of the hockey ball is negative while returning towards its original path.
Recently Updated Pages
Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE General Topics in Chemistry Important Concepts and Tips

Trending doubts
JEE Main 2026: Session 1 Results Out and Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Clemmensen and Wolff Kishner Reductions Explained for JEE & NEET

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding the Angle of Deviation in a Prism

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

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

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

CBSE Notes Class 11 Physics Chapter 14 - Waves - 2025-26

CBSE Notes Class 11 Physics Chapter 9 - Mechanical Properties of Fluids - 2025-26

Inductive Effect and Its Role in Acidic Strength

