Question

A body of mass ${}^{\prime }{m}^{\prime }$ is dropped from a height ${}^{\prime }{h}^{\prime }$. Simultaneously another body of mass $2m$ is thrown up vertically with such a velocity $v$ that they collide at the height $h/2$. If the collision is perfectly inelastic, the velocity at the time of collision with the ground will be :

• A. $\sqrt{\frac{5gh}{4}}$
• B. $\sqrt{gh}$
• C. $\sqrt{\frac{gh}{4}}$
• D. $\frac{\sqrt{10gh}}{3}$

Answer 1

Answer: (D)

$\frac{\sqrt{10gh}}{3}$

First task is to find the initial speed of the body of mass 2m, and time after which the collision occurs. let $v_{0}and{t}_0$ be the initial velocity and time for collision respectively. the kinematic equations are

$\frac{h}{2}=\frac{g{t}_0^2}{2}$and

$\frac{h}{2}=v_0{t}_0-\frac{g{t}_0^2}{2}$

solving for $v_0$ and $t_0$, we get

$t_0=\sqrt{\frac{h}{g}}and{v}_0=\sqrt{gh}$

Next task is to find the velocities of the two bodies just before the collision. let $v_{1}and{v}_2$ be the velocities of bodies of mass m and 2m respectively. kinematic equations are:

$v_1=g{t}_0=\sqrt{gh}\left(downward\right)$
$v_2=v_0-g{t}_0=\sqrt{gh}-g\sqrt{\frac{h}{g}}=0$

Since gravitational force is not impulsive, and time taken for collision is negligible we can conserve the momentum during the perfectly inelastic collision. Thus,

$3m{v}_f=2m\ast 0+m\ast \sqrt{gh}$

$\Rightarrow v_f=\frac{\sqrt{gh}}{3}$

this is the velocity of the the combination of the two bodies just after collsion at height h/2, using the third equation of motion,

$v^2-v_f^2=2g\frac{h}{2}$

putting the values and solving for v,

$\Rightarrow v^2-{\left(\frac{\sqrt{gh}}{3}\right)}^2=gh$
$\Rightarrow v=\frac{\sqrt{10gh}}{3}$