Question

An obiect 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 heighth 20 m. The two point like objects collide and stick to each other. Find the kinetic energy of the combined mass just after the collision.​

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=2×10×5=100.

or v = 10m/s;

Likewise for object A:

by eqn of motion: v = u + at;

wehavev=20−10×1=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

Answer 2

Answer:

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=2×10×5=100.

or v = 10m/s;

Likewise for object A:

by eqn of motion: v = u + at;

wehavev=20−10×1=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

Explanation:

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