Bullet Velocity: Calculate Speed of Bullet Before Hitting Wood

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    Bullet Velocity
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Homework Help Overview

The problem involves a bullet embedding itself into a block of wood on an incline, with the goal of calculating the bullet's speed just before impact. The context includes concepts from mechanics, specifically conservation of momentum and energy.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy and momentum, questioning the correct equations to use before and after the collision. There are attempts to relate potential energy and kinetic energy, as well as clarifications about the roles of mass in the equations.

Discussion Status

Participants are exploring different interpretations of the conservation laws, with some guidance provided on the correct application of equations. There is acknowledgment of confusion regarding the initial and final velocities, and the discussion is actively addressing these points without reaching a consensus.

Contextual Notes

There is mention of a frictionless incline and the need to consider both the bullet and block's masses in the energy and momentum equations. Participants are also navigating through assumptions about initial and final states of the system.

am08
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A bullet of mass m= 3.10×10-2kg is fired along an incline and imbeds itself quickly into a block of wood of mass M= 1.35kg. The block and bullet then slide up the incline, assumed frictionless, and rise a height H= 1.25m before stopping. Calculate the speed of the bullet just before it hits the wood.

What equation do I need?

I was thinking mgh = 1.2mv^2 but that wasn't it.
 
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am08 said:
I was thinking mgh = 1.2mv^2 but that wasn't it.
That's part of what you need. Conservation of energy applies after the collision.

What applies during the collision?
 
am08 said:
AI was thinking mgh = 1.2mv^2 but that wasn't it.

You're nearly there. But you haven't used M, have you?

Try again … :smile:
 
Conservation of Momentum.
 
So i used v2 = squareroot (2gy) and found the final velocity to be 5 m/s

then i used v1 = (mass bullet + mass block / mass bullet) * v2

is this right? I'm using conservation of momentum..
 
Looks good.
 
am08 said:
So i used v2 = squareroot (2gy) and found the final velocity to be 5 m/s

then i used v1 = (mass bullet + mass block / mass bullet) * v2

is this right? I'm using conservation of momentum..

I'm not convinced.

Don't forget - the "final velocity" is zero - you're looking for the initial velocity (the velocity of the bullet just before everything happens).

Your "v = √(2gy)" is wrong - the mass that should be on the left is not the same as on the mass on the right, and so you must put them both in.

(btw, if you type alt-v, it prints √ for you)

Try the √2gy equation (conservation of energy) again! :smile:
 
tiny-tim said:
Your "v = √(2gy)" is wrong - the mass that should be on the left is not the same as on the mass on the right, and so you must put them both in.
Huh? :confused: (Mass cancels out.)
 
Doc Al said:
Huh? :confused: (Mass cancels out.)

No - the inital KE is in the bullet (and the final KE is zero).

The PE involves the block also.
 
  • #10
tiny-tim said:
No - the inital KE is in the bullet (and the final KE is zero).

The PE involves the block also.
Energy is conserved after the collision:

[tex]1/2 (M + m)v_2^2 = (M + m)gy[/tex]

Thus:
[tex]v_2 = \sqrt{2gy}[/tex]
is perfectly correct. (Note that v_2 is the speed of block + bullet immediately after the collision.
 
  • #11
woohooo tiny-tim managed to force the answer out of Doc AI!
 

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