A 25 gm bullet if fired with 3 m/s, hits a block of 2kg, which it at rest, if bullet penetrates the block and moves together in forward , considering elastic impact what will velocity of block after impact.
Answers
conserving the momentum of system.
0.025∗400+4.975∗0.0=(4.975+0.025)∗v
v=10/5=2m/s
this is the answer
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The velocity of block after impact is 0.037 m/s.
A 25 gm bullet if fired with 3 m/s, hits a block of 2kg, which it at rest, if bullet penetrates the block and moves together in forward.
We have to find the velocity of block after impact. (considering elastic impact).
Here external force acting on the system of bullet and block is zero, hence we should apply “law of conservation of linear momentum”
that is, initial linear momentum of system = final linear momentum of system.
⇒mu₁ + Mu₂ = mv + Mv = (m + M)v
- Here, initial velocity of bullet, u₁ = 3 m/s
- initial velocity of block, u₂ = 0 m/s
- final velocity of bullet = v = final velocity of block [ ∵ both moves together so the final velocity must be equal ]
- mass of bullet, m = 25 g = 0.025 kg
- mass of block, M = 2kg
⇒0.025 × 3 + 2 × 0 = (0.025 + 2) × v
⇒v = 0.075/(2.025) m/s ≈ 0.037 m/s
Therefore the velocity of block after impact is 0.037 m/s.
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