A50 gram bullet is fired through a 5 kg block of wood resting on a frictionless surface. the bullet has an initial speed of 500 m/s, while the block is initially at rest. after passing through the block, the bullet has a speed of 300 m/s. how fast is the wooden block moving after the bullet passes through it?

v₂ = 2 m/s

Explanation:

Given that

m₁ = 50 gm , u₁= 500 m/s   ,v₁ = 300 m/s

m₂ = 5 kg   ,u₂=0      

The final speed of the block = v₂

There is no any external force in the system  that is why the linear momentum of the system will be conserved

Initial linear momentum = Final linear momentum

m₁u₁+ m₂ u₂ =m₁v₁+m₂ v₂

0.05 x 500 + 5 x 0 = 0.05 x 300 + 5 x v₂

25 + 0 = 15 + 5 v₂

25 - 15 = 5 v₂

10 = 5 v₂

v₂ = 2 m/s

So the final speed of the wooden block will be 2 m/s.  

total distance/total time

(85 + 82)/(5+3)

167/8

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