A 2.0-kg pistol fires a 1.0-g bullet with a muzzle speed of 1000 m/s. The bullet

Question

A 2.0-kg pistol fires a 1.0-g bullet with a muzzle speed of 1000 m/s. The bullet then strikes a 10-kg wooden block resting on a horizontal frictionless surface. The block and the embedded bullet then slide across the surface.

What is the kinetic energy of the bullet as it travels toward the block?
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I believe it is 500 J based on conservation of linear momentum (is this right ?)
What is the actual math?

Explanation / Answer

its should be mv^2/2 = 1e-3*1000^2/2=500(J)

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