A bullet of mass 10 g moving with a velocity of 400m/s gets embedded in a freely suspended wooden block of mass
900 g. what is the velocity acquired by the wooden block????
plzzzzzzzz tmorrw is my exam do it fast.
Answers
Answered by
235
hope your ans is right and and plz make it brain list
answer:4.4m/s
reason:
let the final velocity of the bullet embedded in it be v
according to the law of conservation of momentum,
total momentum before
collision = total momentum after collision
=
0.001* 400+0.9*0/0.001 + 0.9 = 4/0.91 = 400/91 =
4.4 m/s
answer:4.4m/s
reason:
let the final velocity of the bullet embedded in it be v
according to the law of conservation of momentum,
total momentum before
collision = total momentum after collision
=
0.001* 400+0.9*0/0.001 + 0.9 = 4/0.91 = 400/91 =
4.4 m/s
Answered by
158
By using law of conservation of momentum
m₁u₁ = (m₁ + m₂)v
v = m₁u₁ / (m₁ + m₂)
= (10 × 10^-3 kg × 400 m/s) / [(10 + 900) × 10^-3 kg]
= 4.395 m/s
Velocity acquired by the wooden block is 4.395 m/s
m₁u₁ = (m₁ + m₂)v
v = m₁u₁ / (m₁ + m₂)
= (10 × 10^-3 kg × 400 m/s) / [(10 + 900) × 10^-3 kg]
= 4.395 m/s
Velocity acquired by the wooden block is 4.395 m/s
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