(C) For a perfectly inelastic collision,the loss in kinetic energy $\Delta K$ is given by the formula:$\Delta K = \frac{m_1 m_2}{2(m_1 + m_2)} (u_1 -

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) For a perfectly inelastic collision,the loss in kinetic energy $\Delta K$ is given by the formula:$\Delta K = \frac{m_1 m_2}{2(m_1 + m_2)} (u_1 - u_2)^2$Given:$m_1 = 80 \,kg$,$u_1 = 2 \,ms^{-1}$$m_2 = 20 \,kg$,$u_2 = 4 \,ms^{-1}$Substituting the values:$\Delta K = \frac{80 \times 20}{2(80 + 20)} (2 - 4)^2$$\Delta K = \frac{1600}{2(100)} (-2)^2$$\Delta K = \frac{1600}{200} \times 4$$\Delta K = 8 \times 4 = 32 \,J$Therefore,the loss of energy is $32 \,J$.

Class 11 Physics Work, Energy, Power and Collision Perfectly Inelastic Collision

Medium

An object of mass $80 \,kg$ moving with velocity $2 \,ms^{-1}$ collides with another object of mass $20 \,kg$ moving with velocity $4 \,ms^{-1}$. Find the loss of energy assuming a perfectly inelastic collision. ........... $J$

A
$12$
B
$24$
C
$32$
D
$30$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Work, Energy, Power and Collision

Subtopic

Perfectly Inelastic Collision

$A$ bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ at rest and gets embedded into it. The kinetic energy of the composite block will be

Medium

View Solution

$A$ body of mass $40 \ kg$ moves with a velocity of $4 \ m/s$,and another body of mass $60 \ kg$ moves with a velocity of $2 \ m/s$. If they undergo a perfectly inelastic collision,the loss in kinetic energy is $... \ J$.

Difficult

View Solution

In a one-dimensional collision, a particle of mass $2m$ collides with a particle of mass $m$ at rest. If the particles stick together after the collision, what fraction of the initial kinetic energy is lost in the collision?

Medium

View Solution

$A$ bag of sand of mass $M$ is suspended by a string. $A$ bullet of mass $m$ is fired at it with velocity $v$ and gets embedded into it. The loss of kinetic energy in this process is

Difficult

View Solution

$A$ particle of mass $m$ moving with velocity $v$ collides with a stationary particle of mass $2m$. After collision,they stick together and continue to move together with velocity ........

MediumAIIMS 1999

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

An object of mass $80 \,kg$ moving with velocity $2 \,ms^{-1}$ collides with another object of mass $20 \,kg$ moving with velocity $4 \,ms^{-1}$. Find the loss of energy assuming a perfectly inelastic collision. ........... $J$

Similar Questions

$A$ bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ at rest and gets embedded into it. The kinetic energy of the composite block will be

$A$ body of mass $40 \ kg$ moves with a velocity of $4 \ m/s$,and another body of mass $60 \ kg$ moves with a velocity of $2 \ m/s$. If they undergo a perfectly inelastic collision,the loss in kinetic energy is $... \ J$.

In a one-dimensional collision, a particle of mass $2m$ collides with a particle of mass $m$ at rest. If the particles stick together after the collision, what fraction of the initial kinetic energy is lost in the collision?

$A$ bag of sand of mass $M$ is suspended by a string. $A$ bullet of mass $m$ is fired at it with velocity $v$ and gets embedded into it. The loss of kinetic energy in this process is

$A$ particle of mass $m$ moving with velocity $v$ collides with a stationary particle of mass $2m$. After collision,they stick together and continue to move together with velocity ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module