Body A of mass 4m moving with speed u collide | vedclass

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

Question

The expression of final velocity is given by,

$V _{ B }=\left(\frac{ M _{2}- M _{1}}{ M _{2}+ M _{1}}\right) u _{2}+\frac{2 M _{1} u _{1}}{ M _{2}+

Vedclass · Accepted Answer

$\frac{8}{9}$

Vedclass · Answer

The expression of final velocity is given by,

$V _{ B }=\left(\frac{ M _{2}- M _{1}}{ M _{2}+ M _{1}}\right) u _{2}+\frac{2 M _{1} u _{1}}{ M _{2}+ M _{1}}$

Here,

$M _{1}=4 M , u _{1}= u$

$M _{2}=2 M , u _{2}=0$

Therefore,

$V _{ B }=\left(\frac{2 M -4 M }{2 M +4 M }\right)(0)+\frac{2(4 M ) u }{2 M +4 M }$

$=\frac{4}{3} u$

The fraction of energy lost by colliding body $A$ is as,

$\frac{\frac{1}{2} M_{2} V_{B}^{2}}{\frac{1}{2} M_{1} u_{1}^{2}}=\frac{\frac{1}{2}(2 M)\left(\frac{4}{3}\right)^{2} u^{2}}{\frac{1}{2}(4 M) u^{2}}$

$=\frac{8}{9}$

Body $A$ of mass $4m$ moving with speed $u$ collides with another body $B$ of mass $2 m$ at rest the collision is head on and elastic in nature. After the collision the fraction of energy lost by colliding body $A$ is

Similar Questions

Two bodies with masses $M_1$ and $M_2$ have equal kinetic energies. If $p_1$ and $p_2$ are their respective momenta, then $p_1/p_2$ is equal to

A body constrained to move along $y-$ axis is subjected to a constant force $\vec F = - \hat i + 2\hat j + 3\hat k\,N$ The work done by this force in moving the body a distance of $4\, m$ along $y-$ axis is ............... $\mathrm{J}$

The potential energy of a particle oscillating along $x-$ axis is given as $U =20+ (x - 2)^2$ where $U$ is in $joules$ and $x$ in $meters$ . Total mechanical energy of the particle is $36 \,J$. Maximum kinetic energy of the particle is ............... $\mathrm{J}$

The force $F$ acting on a body moving in a circle of radius $r$ is always perpendicular to the instantaneous velocity $v$. The work done by the force on the body in one complete rotation is :