How Do You Calculate the Final Velocity in a Partially Elastic Collision?

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SUMMARY

The discussion focuses on calculating the final velocity of a bullet after it passes through a block of wood in a partially elastic collision. The bullet has a mass of 4.0 g and an initial speed of 653 m/s, while the block has a mass of 0.093 kg and moves at 24 m/s after the bullet exits. The key equations utilized include conservation of momentum and kinetic energy calculations, specifically using the formulas for initial and final kinetic energy (KE). The correct approach involves calculating the momentum transferred to the block and using it to derive the bullet's final velocity.

PREREQUISITES
  • Understanding of conservation of momentum in collisions
  • Familiarity with kinetic energy calculations (KE = 0.5 * m * v^2)
  • Knowledge of elastic and inelastic collision principles
  • Ability to manipulate algebraic equations for solving physics problems
NEXT STEPS
  • Learn how to apply conservation of momentum in partially elastic collisions
  • Study the differences between elastic and inelastic collisions in detail
  • Practice problems involving kinetic energy calculations before and after collisions
  • Explore advanced collision scenarios, including multi-object interactions
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Students studying physics, particularly those focusing on mechanics and collision theory, as well as educators looking for examples of partially elastic collision problems.

physicsimpaired
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Homework Statement


A bullet with a mass of 4.0 g and a speed of 653 m/s is fired at a block of wood with a mass of 0.093 kg. The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 24 m/s.
(a) What is the speed of the bullet when it exits the block?
wrong check mark
Your answer differs from the correct answer by orders of magnitude. m/s
(b) Is the final kinetic energy of this system equal to, less than, or greater than the initial kinetic energy?
same
less
greater
Explain.
...

After collision speed is reduced for the system and the mass is the same therefore there is less KE than before collision.
(c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system.
Ki =
Kf = wrong check mark Check the syntax of your response.
Your answer differs from the correct answer by 10% to 100%. J

Homework Equations


This is the problem I understand elastic and inelastic separately. but this assumes it is a partially elastic problem and I can't seem to piece together the equation needed from those I have.

Vf = (m1*v1,i + m2v2,i)/(m1 +m2) this assumes though (if I am correct) that the bullet and block become fused.

KE,i = .5*m1*v1^2 + .5*m2*v2^2

KE,f = .5*m1*v1,f^2 + .5*m2*v2,f^2

m1*v,i=m1*v1,f^2 + m2*v2,f

v1,f = ((m1-m2)/(m1+m2))*v1,i

v2,f = ((2*m1)/(m1+m2))v1,i

P = m*v

The Attempt at a Solution


I know for a fact this problem will have some loss in K due to the inelastic portion of the problem. however, I can't find that until i am able to derive the final velocity of the bullet. To answer the homeworks problem in part 5 I need to use the values of velocity for both to find the KE,f.
So the only thing I can't solve is how to find a velocity of an object which collided with and passed through an object.
My attempt was to assume the bullet had "bounced" off in the positive direction. ((.004-.093)/(.004+.093))653 this gave me 599.144 m/s. However this was wrong due to the fact an elastic collision does not reduce KE of the bullet.

I think what I need to do is figure the momentum given to the block by the impact and subtract that from the total initial momentum. This will give me the momentum of the bullet after impact from which I can use the mass to derive the velocity.
 
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physicsimpaired said:
I think what I need to do is figure the momentum given to the block by the impact and subtract that from the total initial momentum. This will give me the momentum of the bullet after impact from which I can use the mass to derive the velocity.
Good! During the collision of bullet and block, momentum is conserved. Use that fact to find the speed of the bullet after it exits the block.
 

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