Question

An object $A$ of mass $1kg$ is projected vertically upward with a speed of $20m/s$. At the same moment another object $B$ of mass $3kg$, which is initially above the object $A$, is dropped from a height $h=20m$. The two point like objects $(A$ and $B)$ collide and stick to each other. The kinetic energy is $K$ (in $J$) of the combined mass just after collision, find the value of $K/25$.

Answer 1

Consider object A attains a height 'x' just before collision and object B descends a height 'y' just before the collision. The collision takes place at time 't' after the two objects start their flight.(Assume 'g' = 10m/s2)

Then for object 'B':

$y=0.5gt²=5m.$

For object 'A':

$x=20t-0.5gt²$

or

$x=20t-y$

or x + y = 20t = 20m;

Hence t = 1 sec.

and x = 15m.

At this time the velocities of objects 'A' and 'B' can be calculated by eqn of motion

v² − u²=2aS

For object B:

$v2=2\times 10\times 5=100.$

or v = 10m/s;

Likewise for object A:

by eqn of motion: v = u + at;

$wehavev=20-10\times 1=10m/s.$

Now by conservation of momentum at the time of collision:

(mA vA - mB vB ) = ( mA + mB )v;

hence

v = mAvA − mBvB/( mA + mB)

or v = -2 x10/4 = -5m/s ie 5m/s in the direction of vB ;

The KE after collision will be

KE = 0.5mAv² + 0.5mBv² = 0.5(mA + mB)v²

or KE = 0.5×4×5² =50J

K/25=2J