A body of mass m moving with velocity v colli | vedclass

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

Question

Velocity of first body after collision is $:-$

$\mathrm{v}_{1}=\left(\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right) \ma

Vedclass · Accepted Answer

$9 : 1$

Vedclass · Answer

Velocity of first body after collision is $:-$

$\mathrm{v}_{1}=\left(\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}ight) \mathrm{u}_{1}+\frac{2 \mathrm{m}_{2} \mathrm{u}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}$

$=\left(\frac{m-2 m}{m+2 m}ight) v+\frac{2 	imes(2 m) 	imes 0}{m+2 m}=\frac{-v}{3}$

$\mathrm{KE}_{\mathrm{i}}=\frac{1}{2} \mathrm{mv}^{2}, \quad \mathrm{KE}_{\mathrm{f}}=\frac{1}{2} \mathrm{m}\left(\frac{-\mathrm{v}}{3}ight)^{2}=\frac{1}{2} \frac{\mathrm{mv}^{2}}{9}$

$\frac{\mathrm{KE}_{i}}{\mathrm{KE}_{\mathrm{f}}}=\frac{\frac{1}{2} \mathrm{mv}^{2}}{\frac{1}{2} \frac{\mathrm{mv}^{2}}{9}}=9$

A body of mass $m$ moving with velocity $v$ collides head on with another body of mass $2\, m$ which is initially at rest. The ratio of $K.E.$ of the colliding body before and after collision will be

Similar Questions

A vertical spring with force constant $K$ is fixed on a table. A ball of mass $m$ at a height $h$ above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance $d$. The net work done in the process is

A block of mass $0.50\, kg$ is moving with a speed of $2.00\, ms^{-1}$ on a smooth surface. It strikes another mass of $1.00\, kg$ and then they move together as a single body. The energy loss during the collision is .............. $\mathrm{J}$

If the potential energy of a gas molecule is

$U = \frac{M}{{{r^6}}} - \frac{N}{{{r^{12}}}}$,

$M$ and $N$ being positive constants, then the potential energy at equilibrium must be

$A$ ball is dropped from height $5m$. The time after which ball stops rebounding if coefficient of restitution between ball and ground $e = 1/2$, is .................. $\mathrm{sec}$

Pulley and spring are massless and the friction is absent everwhere. $5\,kg$ block is released from rest. The speed of $5\,kg$ block when $2\,kg$ block leaves the contact with ground is (take force constant of the spring $K = 40\,N/m$ and $g = 10\,m/s^2$ )