A 10g bullet travelling at 20m/s strikes and remains embedded in a 2kg target which is originally at rest but free to move. At what speed does the target move off.
Options
1
0.1 m/s
2
less than 0.1 m/s
3
greater than 0.1 m/s
4
data insufficient
Answers
Answer:
A 10g bullet travelling at 200m/s strikes and remains embedded in a 2kg target which is originally at rest but free to move.At what speed does the target move off?
Posted by Khushi Joshi 1 year, 10 months ago
CBSE > Class 09 > Science
1 answers
Govind Singh 1 year, 5 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
∴v=22−01
Explanation:
Hence, the target will move off with the speed of 0.99 m/s
Answer:
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