Momentum - Trend of the day

[HollowSuperet36]

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The other problem that didn't click.

Two billiard balls undergo a head-on collision. Ball 1 is twice as heavy as ball 2. Initially, ball 1 moves with a speed V towards ball 2 which is at rest. Immediately after the collision, ball 1 travels at a speed of V/3 in the same direction. What type of collision has occured?

A, Inelastic
B. Elastic
C. Completely Inelastic
D. Cannot be determined from the information given.

B. Elastic

 

[IRASNA]

I believe it's completely in elastic meaning they have stuck together.

mv= (m + 2m)(new v)
mv=(3m)(new v)
New V must be V/3

Correct me if I'm wrong.

 

[railgun]

m1v1=m1v2+m2v3 => m1v1=m1v2+2m1v3=> v1=v2+2v3
v1 is ball 1, v2 is ball 1 after collision, v3 is ball 2 after collision
so plug in:
V=V/3+2V3=> V3=1/3V
KE:
1/2mV^2 vs 1/2*2m*V^2/9+1/2*m*V^2/9=> KE not conserved
so because Both balls are moving at V/3 in same direction it is a completely inelastic collision.
C
Edit:

New V must be 3

Correct me if I'm wrong.

It should be V/3
Edit again:
I'm an idiot. I used ball 2 weighing 2x as ball 1
so new answer:
2*m1v1=2*m1v2+m1v3
v1 is ball 1, v2 is ball 1 after collision, v3 is ball 2 after collision
so plug in:
2V=2V/3+V3=> V3=4V/3

 

[milski]

Momentum before the collision is 2mv, energy is mv^2 (all from ball with velocity v and mass 2m)

After collision the momentum is 2mv/3+mx. Momentum should stay the same, solve for x and you get 4v/3 for velocity of the small ball. Calculate the total energy of both balls now and it ends up the same. No energy loss - the collision is elastic.


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[IRASNA]

m1v1=m1v2+m2v3 => m1v1=m1v2+2m1v3=> v1=v2+2v3
v1 is ball 1, v2 is ball 1 after collision, v3 is ball 2 after collision
so plug in:
V=V/3+2V3=> V3=1/3V
KE:
1/2mV^2 vs 1/2*2m*V^2/9+1/2*m*V^2/9=> KE not conserved
so because Both balls are moving at V/3 in same direction it is a completely inelastic collision.
C
Edit:


It should be V/3

Mobile SDN'ing while walking down the stairs....innocent mistake. Thanks

 

[milski]

m1v1=m1v2+m2v3 => m1v1=m1v2+2m1v3=> v1=v2+2v3
v1 is ball 1, v2 is ball 1 after collision, v3 is ball 2 after collision
so plug in:
V=V/3+2V3=> V3=1/3V
KE:
1/2mV^2 vs 1/2*2m*V^2/9+1/2*m*V^2/9=> KE not conserved
so because Both balls are moving at V/3 in same direction it is a completely inelastic collision.
C
Edit:


It should be V/3

I think the first ball is 2m?


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[railgun]

I think the first ball is 2m?


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Yea, you got it right. I hate these stupid mistakes. I'm just glad i didn't make too many on my mcat!

 

[johnwandering]

..

 

[Girly X]

Yes elastic is right because mechanical energy is conserved. In other words ball 1 started with 1v then ended with (1/3)v meaning (4/3)v was " lost" but it was not lost because all of it was transferred to ball 2. Solve for x in 2m.v + 0 = 2m.v/3 + m.x. You will find that x=4/3