Energy Conserved in momentum problem

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SUMMARY

The discussion centers on an elastic collision problem involving two 4.0 kg bowling balls, where one ball, initially moving at 8.0 m/s, collides head-on with a stationary ball. After the collision, the first ball stops, and the second ball moves at 8.0 m/s. The analysis confirms that kinetic energy is conserved, demonstrating that the collision is perfectly elastic. The relevant equations used include momentum conservation (m1v1 + m2v2 = m1v1' + m2v2') and kinetic energy calculations (KE = 0.5mv²).

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Homework Statement


A 4.0kg bowling ball rolling to the right at 8.0 m/s has an elastic head-on collision with another 4.0kg ball initially at rest. The first ball stops after the collision.
a) Find the velocity of the canoe after the collision?
b) Is the collision perfectly elastic? (show work)
c) Is energy conserved? (show work)

Homework Equations


m1v1+m2v2 = m1v1' + m2v2'
PE=KE
mgh=1/2mv2

The Attempt at a Solution


a) (4.0kg)(8.0m/s) + (4.0kg)(0 m/s) = (4.0kg)(0 m/s) + (4.0kg)(v2')
32 kgm/s + 0 kgm/s = 0 kgm/s + 4.0kg(v2')
8 m/s = v2'

b) Don't know how to tell if it is perfectly elastic, no equation that I can think of

c) Don't know how to tell if energy is conserved, no equation that I can think of
 
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What equations do you know of that deals with energy?
 
I know KE = PE, and that is 1/2 mv2 = mgh
 
So we know that in an elastic collision, energy is conserved, right? So is there a way to compare the before and after collision energy?
 
I do know that no energy is lost, so that means energy is conserved. As to a way to compare the before and after collision energy..I'm sifting through my notes and not seeing anything glaring out.
 
Well, you could calculate the energy of the system before and after, right?

Hint: Kinetic energy.
 
.5(4.0kg)(8.0m/s)2 = 128 J = .5(4.0kg)(8.0 m/s)2

oh...so the KE is the same on both sides, so no energy is lost.

and if no energy is lost, then it has to be perfectly elastic?
 
Yup, simple as that!
 
Woopy said:

Homework Statement


A 4.0kg bowling ball rolling to the right at 8.0 m/s has an elastic head-on collision with another 4.0kg ball initially at rest. The first ball stops after the collision.

Homework Equations


m1v1+m2v2 = m1v1' + m2v2'
PE=KE
mgh=1/2mv2
You should also be aware that PE has nothing to do with this problem. Typically, you don't consider PE in collision problems between hard particles (eg. bowling balls), gravitational or otherwise. An exception would be if the balls were rolling on a slanted surface, or if a ball collides with a spring, but usually you just ignore PE in a collision problem.
 

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