How High Will the Bullet-Carrying Cart Climb the Ramp?

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SUMMARY

The discussion focuses on calculating the maximum height a bullet-carrying cart will reach after being fired into a wood block mounted on a low-friction cart. A 15 g bullet traveling at 150 m/s embeds itself in a 0.5 kg cart, resulting in a combined velocity of 4.37 m/s post-impact. The cart ascends a 30-degree ramp, and using energy conservation principles, the maximum height is calculated to be approximately 1.2 meters. The participants emphasize the importance of applying the correct physics equations, particularly the conservation of momentum and energy.

PREREQUISITES
  • Understanding of conservation of momentum
  • Familiarity with energy conservation principles
  • Knowledge of basic kinematics
  • Ability to perform trigonometric calculations for angles
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  • Study the principles of conservation of momentum in inelastic collisions
  • Learn about energy conservation in mechanical systems
  • Explore kinematic equations for motion on inclined planes
  • Investigate the effects of friction on motion and energy loss
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of collisions and motion on inclines, particularly in experimental settings.

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Homework Statement



At a crime lab, they are experimenting measuring the speed of bullets. In it, a 15 g bullet is fired at 150 m/s into a wood block mounted on a low-friction cart (m=0.5kg). The bullet imbeds in the cart and the cart then rolls up a 30-degree ramp. Ignoring friction, predict the max height that the cart will rise to as it rolls up the lamp.

Homework Equations





The Attempt at a Solution



.015(150)+0=(.5+.015)vx

vx=4.37 ms

since it's a 30 degree angle.. the velocity of cart=4.37/cos(30)=5.04 m/s

x=(vf^2-vo^2)/2a
=(5.04^2-0)/2(9.8)
-1.2?

i don't know if this was correct.
 
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Why not just say mgh=.5m(vx)^2

h = sqrt(4.37^2/20)
 

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