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miskamm [114]
2 years ago
14

A 0.11 kg bullet traveling at speed hits a 18.3 kg block of wood and stays in the wood. The block with the bullet imbedded in it

moves forward with a velocity of 8.8 m/s. What was the velocity (speed) of the bullet immediately before it hit the block (in m/s)?
Physics
1 answer:
arlik [135]2 years ago
5 0

Explanation:

The energy of the system before the collision must equal the energy after the collision.

After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is

\frac{18.41 kg (8.8 m/s)^2}{2} = 713 J

Before the collision only the bullet has kinetic energy.

So we can now determine the speed of the bullet using

\frac{0.11kg (v^2)}{2} = 713 J\\v = 114 m/s

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<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
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