JEE Class main Answered
As shown in Figure, a bullet weighs 0.3 N and moving with a velocity of 400 m/sec hits centrally a 30 N block of wood moving away at 15 m/sec and gets embedded in it. Find the velocity of the bullet after the impact and the amount of kinetic energy lost.
Asked by theavengers0203 | 22 May, 2023, 11:54: PM
Expert Answer
Mass of bullet = 0.3 / 9.8 = 0.031 kg
Mass of wooden block = 30 / 9.8 = 3.06 kg
Initial momentum of bullet = mass × velocity = 0.031 × 400 = 12.4 kg m/s
Initial momentum of wooden block = 3.06 × 15 = 45.9 kg m/s
Total Initial momentum = ( 12.4 + 45.9 ) kg m/s = 58.3 kg m/s
Bullet is embedded in the block after hitting the block. Hence collision is perfect inelstic collision.
Only Momentum is conserved.
Final velocity of block embedded with bullet = momentum / mass
= 58.3 / ( 3.06 + 0.031 ) = 18.86 m/s
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Initial kinetic energy of bullet = (1/2) m v2 = 0.5 * 0.031* 400 * 400 J = 2480 J
Initial kinetic energy of wooden block = (1/2) M vB2 = 0.5 * 3.06 * 15 * 15 J = 344 J
Total initial kinetic energy = ( 2480 + 344 ) J = 2824 J
Final kinetic energy of wooden block and bullet = (1/2) ( M +m) V2
= 0.5 × ( 3.06 + 0.031 ) * 18.86 × 18.86 J = 550 J
Loss of kinetic energy = ( 2824 - 550 ) J = 2274 J
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