Question

A lead bullet of mass 50 grams is fired with a velocity of 200m/s into a lead block of mass 950 grams. Given that the lead block can move freely, calculate the kinetic energy after impact and the loss of energy.

Answer 1

Convert the masses in g to kg
So, 50g=0.05kg & 950g=0.95kg
Mass of bullet (Mb)=0.05kg
Mass of lead block (mB)=0.95kg
Initial vel of bullet (Ub)=200m/s
Final vel of bullet (Vb)=0m/s
Initial vel of lead block(uB)=0m/s
Final vel of lead block(vB)=vB?
The collision is linear. So, using: 1/2MbUb^2+1/2mBuB^2=1/2MbVb^2+1/2mBvB^2
1/2×0.05×200^2+1/2×0.95×0^2=1/2×0.05×0^2+1/2×0.95×vB^2
1000+0=0+0.5vB^2
1000=0.5vB^2
Therefore VB^2=1000/0.5
Hence vB=√2000
vB=45m/s (approximately)
After impact, they will move with the same velocity i.e vel of lead block
Therefore, Kinetic Energy after impact =kinetic energy of lead block=k.eB
k.eB=1/2mBvB^2
k.eB=1/2×0.95×45^2
k.eB=962Joules(approximately)
Recall; Mb=0.05kg
Ub=200m/s
Therefore, K.Eb(kinetic energy of bullet)
K.Eb=1/2MbUb^2
K.Eb=1/2×0.05×200^2
K.Eb=1000Joules
LOSS OF ENERGY =K.Eb-k.eB
=1000-962=38Joules

Answer 2

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