Two bullets of mass 250 g moving in opposite directions with a speed of 500 m/s and 250 m/s collide with a stationary block of mass 12 kg. If the bullets stick to the block and all of them move together, what is the the final velocity of the system?

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Solution

For bullet 1(writing as B1)
initial momentum, p1= mv=250×10^-3 ×(500)=125 kgm/s.
For B2,
Initil momentam= -P2 (B2 is moving opposite to B1)

mv=250×10^-3 ×(250)=62.5 kgm/s.

Initial momentum of the block,P3=0.
Mass of block=12kg.

total intial momentum of the system=P1+ -P2+P3. =125+ -62.5 +0=62.5 kgm/s.

Final momentum will be=M×V.
Final mass of the system,M=m1+m2+m3=(250+250)×10^-3 kg + 12kg=12+.5=12.5 kg

[ 1g=10^-3 kg ]

By conservation of momentum => intial mometum of the system=final momentum of the system.
from the above, final velocity= initial momentum of system/ (final mass of the system).
That is,
V= 62.5 / (12.5)= 5m/s.
Total system will move with 5m/s in the direction of B1.(because B1 moves with higher velocity than B2).

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