Question

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

• A. $\sqrt{\frac{5gh}{4}}$
• B. $\sqrt{gh}$
• C. $\sqrt{\frac{gh}{4}}$
• D. None of these

Answer 1

Answer: (D)

None of these

m mass falls a distance $\frac{h}{2}$ in time, say t. Then,
$\frac{h}{2}=\frac{1}{2}g{t}^2\therefore t=\sqrt{\frac{h}{g}}$
For mass 2m,
$\frac{h}{2}=vt-\frac{1}{2}g{t}^2=v\sqrt{\frac{h}{g}}-\frac{h}{2}$
$\therefore v=\sqrt{gh}$
Now, let u is the velocity with which combined mass collides with ground, then
$u=\sqrt{(\frac{2}{3}v{)}^2+2g(\frac{h}{3})}$
$=\sqrt{\frac{4}{9}gh+\frac{2}{3}gh}$
$=\frac{\sqrt{10}}{3}\sqrt{gh}$

Since the collision takes place in vertical direction and external force is not zero in this direction, momentum conservation cannot be applied.
If the gravitational force( weight) is much less than the impulse force, it may be ignored and conservation of momentum can be applied.