A 10-g bullet with an initial speed of 292 m/s is shot directly at a 1-kg wooden

Question

A 10-g bullet with an initial speed of 292 m/s is shot directly at a 1-kg wooden block, which rests on a frictionless surface. What is the speed of the block, after the bullet is embedded in the block?

(Although this is not justified, calculate your answer to 4 significant figures. This will uncover any conceptual errors you might make.)

I got 2.92m/s but that was wrong not sure if I imputed the wrong values

A 10-g bullet with an initial speed of 292 m/s is shot directly at a 1-kg wooden block, which rests on a frictionless surface. What is the speed of the block, after the bullet is embedded in the block?

(Although this is not justified, calculate your answer to 4 significant figures. This will uncover any conceptual errors you might make.)

I got 2.92m/s but that was wrong not sure if I imputed the wrong values

Explanation / Answer

The key point here is bullet and wooden block move together after collision

mass of the bullet m = 10 g = 10 * 10-3 kg = 0.01 kg

intial velocity of the bullet v = 292 m/s

mass of the block M = 1 kg

intial velocity of block u = 0

Common velocity of the bullet and Wooden block be V

So equatin for conservation of momentum is

mv + Mu = (m + M)V

0.01 * 292 + 0 = (0.01 + 1)V

V = 2.92/1.01

= 2.89109 m/s

so velocity of the wooden block is 2.89109 m/s

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