Momentum and Energy related Bullet Question

AI Thread Summary
The discussion revolves around calculating the initial speed of a bullet that strikes a block, leading to a perfectly inelastic collision. Participants suggest using conservation of momentum to find the combined velocity of the bullet and block after the collision. The angle of deflection is important for determining the height gained by the block, which can be calculated using trigonometry. This height is then used to apply conservation of energy principles, equating kinetic energy to gravitational potential energy. The correct approach involves combining these concepts to solve for the bullet's initial velocity.
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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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