Question

A 9.7-g bullet is fired into a stationary block of wood having mass m = 5.02 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.610 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.)

Answer 1

Given :

Mass of bullet = M_{​​​​​​1} = 9.7 * 10^-3 kg

Mass of block of wood = m = 5.02 kg

Let initial speed of bullet = V_{​​​​​​1} m/s

Final velocity of bullet + wood = 0.610 m/s

Now applying conservation of momentum :-

Initial momentum of bullet + wood = finlal momentum of bullet+ wood

M_{​​​​​​1}* V_{​​​​​​1} + m* v = V_{​​​​​​f} ( M_{​​​​​​1} + m)  

9.7 * 10^-3 * V_{​​​​​​1} + 0 = 0.610 * ( 9.7 * 10^-3 + 5.02 ) ( v= 0 , as

wood is at rest)

V_{​​​​​​1} = 316.30 m/s

Original speed of bullet = V_{​​​​​​1}= 316.30 m/s