What Is the Initial Velocity of the Bullet?

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The discussion revolves around calculating the initial velocity of a bullet fired into a wooden block, along with the block's velocity post-impact and the energy lost during the collision. The bullet's mass is 0.007 kg, and it exits the block at 200 m/s, while the block, initially at rest, slides 0.5 m before stopping. The initial velocity of the bullet was calculated to be approximately 202.09 m/s using the work-energy theorem, while the block's velocity right after the impact was found to be 1.714 m/s. Energy loss during the collision was estimated at 2.94, although there was some confusion regarding the calculations for parts a and c. The conversation highlights the importance of using conservation of momentum for accurate results in inelastic collisions.
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Homework Statement


A .007 kg bullet is fired into a 2 kg wooden block initially at rest on a level surface. The bullet passes through the block and emerges with the velocity of 200 m/s. The block slides .5 m and stops. If the coefficient of the friction between the block and and the surface is .3, find a) the initial velocity of the bullet, b) the velocity of the block right after being hit by the bullet and c) the energy loss of the bullet during the collision.

Homework Equations


I understand how to get part a of the question but I can't figure out the other two parts. Please help me :[.

The Attempt at a Solution


For part a I got 202.09 m/s.
 
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Hi PreetiM! Welcome to PF! :wink:

Use the work energy theorem …

work done = change in kinetic energy …

show us what you get. :smile:​
 
Hello! Thanks for helping :)

So for part b) I used Fd= 1/2mvi2-1/2mvf2
and for the velocity of the block I got 1.714 m/s.

For part c) I just subtracted the final kinetic energy from the initial kinetic energy and got 2.94. I think that looks correct :P.
 
(b) looks correct. :smile:

I don't understand how you got (a) or (c) :confused: …

can you show your calculations?​
 
For a) I did what a peer tutor told me to do, he said
Initial Kinetic energy - work done by friction = final kinetic energy.

So I did, .5mvi2 - Ffrictiond = .5mvf2

When I solved for vi I got 202.089 m/s.

I admit for part c) I did just guess :(.
 
Either your peer tutor is wrong, or he was talking about part (b) not part (a). :redface:

For part (a), energy is not conserved (this is obviously not an elastic collision ).

Use conservation of momentum instead (you'll need the result from part (b) for this).
 

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