Momentum and Energy related Bullet Question

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SUMMARY

The discussion focuses on calculating the initial velocity of a 5.00 g bullet that is fired into a 6.00 kg block suspended from a 1.00 m string, which deflects at an angle of 12.0°. The key approach involves using the conservation of momentum to find the final velocity after the perfectly inelastic collision and then applying conservation of energy principles to relate kinetic energy to gravitational potential energy. The angle of deflection is utilized to determine the height change, which is essential for calculating the energy transformation.

PREREQUISITES
  • Understanding of conservation of momentum in perfectly inelastic collisions
  • Knowledge of kinetic energy and gravitational potential energy equations
  • Familiarity with trigonometric functions, specifically cosine for height calculations
  • Basic principles of mechanics related to collisions and energy transformations
NEXT STEPS
  • Study the conservation of momentum in detail, focusing on perfectly inelastic collisions
  • Learn how to apply conservation of energy in mechanical systems
  • Explore trigonometry applications in physics, particularly in calculating heights from angles
  • Review examples of energy transformations in collision scenarios
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of momentum and energy in collision problems.

tmang
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A 5.00 g bullet is fired into a 6.00 kg block, which is suspended from a string 1.00 m long.
The string deflects through an angle of 12.0°. How fast was the bullet moving?a




Transitional KE = ½ mv2

Rotational KE = ½ I2 Elastic PE = ½k L 2

I'm not sure those are the right formulas even...




I'm not really sure what to do with the angle..For the left of the equation, I know that the first mass is the bullet's mass and the block's mass, but the block's mass will be ignored because it was not moving originally. For the right of the equation, the mass will be combined and the final velocity will be... I think the angle has something to do with the final velocity but somehow I can't figure out in any way how to deal with it.


Thank you for your help.. please help me!
 
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You have perfectly inelastic collision. It's easy to get the velocity of the block and the bullet togehter using conservation of momentum. Then you can use conservation of energy (kinetic energy is equal to change of gravitational potential energy). To calculate that change, you need the height on which bullet has climbed. That's why you have the angle and the length of the string. It's simple trigonometry (cosine in this case).
 
tmang said:
A 5.00 g bullet is fired into a 6.00 kg block, which is suspended from a string 1.00 m long.
The string deflects through an angle of 12.0°. How fast was the bullet moving?a




Transitional KE = ½ mv2

Rotational KE = ½ I2 Elastic PE = ½k L 2

I'm not sure those are the right formulas even...




I'm not really sure what to do with the angle..For the left of the equation, I know that the first mass is the bullet's mass and the block's mass, but the block's mass will be ignored because it was not moving originally. For the right of the equation, the mass will be combined and the final velocity will be... I think the angle has something to do with the final velocity but somehow I can't figure out in any way how to deal with it.


Thank you for your help.. please help me!

First solve using the conservation of momentum equation: m1v1 +m2v2 = (m1 + m2)vf
 

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