Question

A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its stand. what was the final velocity of the bullet?

Answer 1

Answer:

The final velocity of the bullet is 9 m/s.

Explanation:

We have,

Mass of a bullet is, m = 0.05 kg

Mass of wooden block is, M = 5 kg

Initial speed of bullet, v = 909 m/s

The bullet embeds itself in the block which flies off its stand. Let V is the final velocity of the bullet. The this case, momentum of the system remains conserved. So,

$mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s$

So, the final velocity of the bullet is 9 m/s.