Calculating Bullet Speed in Inclined Slide | Mass, Height & Friction Considered

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Homework Help Overview

The problem involves calculating the speed of a bullet just before it embeds into a block of wood on an inclined plane, considering the mass of the bullet and block, the height reached after the collision, and the assumption of a frictionless incline.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of conservation of momentum and energy, with questions about the necessary parameters such as the angle of the incline and the final velocity of the block and bullet system.

Discussion Status

Some participants have offered guidance on using conservation of energy to find the required velocity for the block and bullet to reach a certain height. There is an acknowledgment of a misunderstanding regarding the application of energy conservation during the impact.

Contextual Notes

Participants are navigating assumptions about the nature of the collision (elastic vs. inelastic) and the implications for energy conservation in their calculations.

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A bullet of mass m= 0.0260 kg is fired along an incline and imbeds itself quickly into a block of wood of mass M= 1.30 kg. The block and bullet then slide up the incline, assumed frictionless, and rise a height H= 1.20 m before stopping. Calculate the speed of the bullet just before it hits the wood. Note. The block is kept from sliding down the incline initially by as small peg (not shown).
I know that I need to use conservation of momentum, but I'm not sure how I would do that since I don't know the final velocity. I don't even know how to find it since I don't know the angle of the incline of the length of the incline.
Any advice?
Thanks
 
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What is the velocity required for the block (with the embedded bullet) to reach a height of 1,20 m?
 
Start by thinking about conservation of energy.
What kind of eqn(s) are/is involved in doing that ?
 
Ok for conservation of energy
I used (1/2)mv_initial^2=Mgh
(1/2)(.0260)v^2= (.0260+1.30)(9.8)(1.20)
Solving for v gave me 34.6 m/s.
Am I doing that right?
 
No, you are not. You cannot assume energy conservation during impact (unless it is stated that the collision is elastic, which is not the case in this problem).

After impact, however, energy is conserved.
 
I figured out what I was doing wrong. Thank you.
 

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