A bullet weight 15g and travelling at 500m/s collides with an 0.8kg block at res

Question

A bullet weight 15g and travelling at 500m/s collides with an 0.8kg block at rest on the end of a table

If the collision is perfectly inelastic find

a: Determin the velocity of the block as it leaves the table and how far from the end of the table will the block hit the floor

b: Determine the mechanical energy that disappears as a result of the collision. Account for this missing energy.

I have worked out the part A which the velocity is 9.2m/s and distance is 3.72m

Can I please have help with part B? thanks

Explanation / Answer

Here Intial Kinetic Energy = 0.5*15*10^-3*500^2 = 1875 J As you said that (Block+bullet) velocity become 9.2 m/sec after Collision Therefore Final Kinetic Energy = 0.5*(92*10^-3 + 0.8)*9.2^2 = 37.75 J Therefore Loss in Kinetic Energy = 1875 - 37.75 = 1837.25 J This is the Mechanical Energy that Disapper

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