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Question

An object A of mass 1 kg is projected vertically upward with a speed of 20 m/s. At the same moment another object B of mass 3 kg, which is initially above the object A, is dropped from a height h=20 m. 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.

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Solution

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=20t0.5gt²
or
x=20ty
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=2×10×5=100.
or v = 10m/s;
Likewise for object A:
by eqn of motion: v = u + at;
wehavev=2010×1=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

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