A 10g bullet travelling at 200m/s …
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Govind Singh 7 years, 11 months ago
Total momentum before collision
= (mass of the bullet x velocity) + (mass of the target x velocity)
= (0.01kg x 200m/s) + 2kg x 0)
= 2kg m/s
Total momentum after collision
= (mass of the bullet + mass of the target) x velocity
= (0.01kg + 2kg) x v
According to the law of conservation of momentum.
Total momentum before collision = Total momentum after collision
Total momentum before collision = Total momentum after collision
= 2kg /s = (2.01)v
{tex}\therefore v = \frac{2}{{2 - 01}}{/tex}
Hence, the target will move off with the speed of 0.99 m/s
1Thank You