Question

A 0.01 kg bullet has a speed of 700m/s before it strikes a 0.95 kg wooden block that is stationary on a horizontal frictionless surface and remains inside of it. What is the speed of the block after the bullet becomes embedded in it?

Answer 1

The final speed is 7.29 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum: in fact, the total momentum of the bullet-block system must be conserved before and after the collision. Therefore, we can write:

$p_i = p_f\\mu=(m+M)v$

where:

m = 0.01 kg is the mass of the bullet

u = 700 m/s is the initial velocity of the bullet

M = 0.95 kg is the mass of the block

v is the final combined velocity of bullet+block

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

$v=\frac{mu}{m+M}=\frac{(0.01)(700)}{0.95+0.01}=7.29 m/s$

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