Momentum and Energy related Bullet Question

In summary: Plug in values and solve for vf:0.005kg * v1 + 6.00kg * 0 = (0.005kg + 6.00kg) * vf0.005kg * v1 = 6.005kg * vfvf = 0.005kg * v1 / 6.005kgNext, use
  • #1
tmang
3
0
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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  • #2
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).
 
  • #3
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
 

What is momentum?

Momentum is a measure of an object's motion, determined by its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.

How is momentum related to energy?

Momentum and energy are closely related. Momentum is a conserved quantity, meaning it cannot be created or destroyed. This is similar to the conservation of energy, where energy can change forms but cannot be created or destroyed.

What is the difference between kinetic and potential energy?

Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object has due to its position or state. In the context of a bullet, kinetic energy is the energy it has while in motion, and potential energy would be the energy it has due to its position in the gun before being fired.

Why is it important to consider momentum and energy when firing a bullet?

Momentum and energy play a crucial role in the trajectory and impact of a bullet. The momentum of the bullet can determine its recoil and the force it exerts on its target. The energy of the bullet can determine its velocity and penetration power.

How can momentum and energy be conserved in a bullet's trajectory?

Momentum and energy can be conserved in a bullet's trajectory through the principles of Newton's laws of motion. This means that the initial momentum and energy of the bullet must be equal to the final momentum and energy after it has been fired. Factors such as air resistance and friction may cause some loss of momentum and energy, but overall they should remain conserved.

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