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?

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:

[tex]p_i = p_f\\mu=(m+M)v[/tex]

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:

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

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pstnonsonjoku pstnonsonjoku · Answer 2 · 2022-04-05T14:05:52+02:00

The velocity of the block after the bullet becomes embedded in it is 7.29 m/s.

What is momentum?

The term momentum is the product of mass and velocity. We know that from the principle of conservation of linear momentum; momentum before collsion is equal to momentum after collsion.

Given that;

(0.01 kg * 700m/s) + (0.95 kg * 0 m/s) = (0.01 kg + 0.95 kg) v

v = 7.29 m/s

The velocity of the block after the bullet becomes embedded in it is 7.29 m/s.

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