Question

A 0.001kg bullet is fired with a velocity of 800m/s into a soft wood of mass 1kg resting on a smooth surface. Find the final velocity if the bullet is fully embedded I n the wood after the collision​

Answer 1

The final velocity of the bullet+block is 0.799 m/s

Explanation:

We can solve this problem by applying the principle of conservation of momentum: in fact, the total momentum of the bullet-block system must be conserved before and after the collision.

Mathematically, we can write:

$mu+MU=(m+M)v$

where

m = 0.001 kg is the mass of the bullet

u = 800 m/s is the initial velocity of the bullet

M = 1 kg is the mass of the block

U = 0 is the initial velocity of the block (initially at rest)

v is the final combined velocity of the bullet and the block

Solving the equation for v, we  find the final velocity:

$v=\frac{mu}{m+M}=\frac{(0.001)(800)}{0.001+1}=0.799 m/s$

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