Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

In a collinear collision, a particle with an initial speed vo strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :

Option 1)
$\frac{V_{0}}{\sqrt{2}}$

Option 2)
$\frac{V_{0}}4$

Option 3)
$\sqrt{2} v_{0}$

Option 4)
$\frac{V_{0}}2$

Answers (2)

As we learnt that

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$m_{1},m_{2}:masses$

$u_{1},v_{1}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{1}$

$u_{2},v_{2}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{2}$

From conservation of momentum

$mv_{o}+0=mv_{1}+mv_{2}\; \; \; \; \; \; eq \, 1$

and $\frac{3}{2}(\frac{1}{2}mv^{2})=\frac{1}{2}mv_{1}^{2}+\frac{1}{2}mv_{2}^{2}$

$or\: v_{1}^{2}+v_{2}^{2}=\frac{3}{2}v_{o}^{2}$ eq 2

^{From 1 and 2}

Option 1)

$\frac{V_{0}}{\sqrt{2}}$

This is incorrect

Option 2)

$\frac{V_{0}}4$

This is incorrect

Option 3)

$\sqrt{2} v_{0}$

This is correct

Option 4)

$\frac{V_{0}}2$

This is incorrect

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today

anam-khan

BTech admission 2024 without JEE; top government engineering colleges, upcoming entrance exams

anam-khan

JEE Main 2024 answer key out at jeemain.nta.ac.in; raise objections till April 14

Latest Question