Question 26:
A 10 g bullet travelling at 200 m/s strikes and remains embedded in a 2 kg target which is originally at rest but free to move. At what speed does the target move off ?
Lakhmir Singh Physics Class 9
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the formula of the class 9th is l into S is equal to be the former is the formula is array of constant in health 2081 competition and everyone around you are so big around in our world 9th class of the over the formula is equal to b at thank
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Answer:
0.99m/s
Explanation:
Mass of the bullet = m1 = 10g = 0.01 kg (Given)
Velocity of the bullet = v1 = 200 m/s (Given)
Let the recoil velocity be = v2
Mass of thr block when the bullet gets embedded in it will be -
m2 = 2 +0.01
= 2.01 kg
According to the law of conservation of momentum -
m1 × v1 = m2 × v2 0.01 × 200
= 2.01 × v2
v2 = 0.01 × 200/2.01
v2 = 0.99
Thus, speed does the target will move off at 0.99m/s
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