Elastic Collision of Two Objects (2D)

AI Thread Summary
The discussion revolves around solving a physics problem involving an elastic collision between two objects with different masses and velocities. The user initially struggles with the equations for conservation of momentum and kinetic energy, attempting to express one variable in terms of another. Clarifications are provided regarding the correct formulation of the energy conservation equation, emphasizing the need for two equations to solve for the two unknown velocities. Additionally, a relationship for relative velocity in elastic collisions is introduced as a quicker method to find the final velocities. The conversation highlights the importance of correctly applying physics principles to arrive at the solution.
Hermit Solmu
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Homework Statement



A 17g object moving to the right at 32 cm/s overtakes and collides elastically with a 35g object moving in the same direction at 12 cm/s. Find the velocity of the slower object after the collision. Answer in cm/s.

B) Find the velocity of the faster object after the collision. Answer in cm/s.


Homework Equations



(ma1*va1)+(mb1*vb1)=(ma2*va2)+(mb2*mb2)

.5((ma1*va1)^2+(mb1*vb1)^2)=.5((ma2*va2)^2+(mb2*mb2)^2)

The Attempt at a Solution



(17*32)+(35*12)=(17*x)+(35*y)

And since I have two variables, I'm not sure how to solve for either of them. I tried solving for one in terms of the other...

(928-17x)/35=y

But that wasn't the answer my instructor was looking for. Thanks so much for any help you can provide.
 
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Hermit Solmu said:
(ma1*va1)+(mb1*vb1)=(ma2*va2)+(mb2*mb2)
OK.

.5((ma1*va1)^2+(mb1*vb1)^2)=.5((ma2*va2)^2+(mb2*mb2)^2)
You have typos in this one. Each term should be ½mv², not ½(mv)².

The Attempt at a Solution



(17*32)+(35*12)=(17*x)+(35*y)
So far, so good.

And since I have two variables, I'm not sure how to solve for either of them. I tried solving for one in terms of the other...

(928-17x)/35=y
Now use your 2nd equation, the one for energy conservation. (2 unknowns requires 2 equations. Luckily you have 2 equations.)
 
So you're saying my second equation should be .5m^2+.5v^2=.5m2^2+.5v2^2?

And then just set the Y equal to each other, then do the same thing with the X?
 
Hermit Solmu said:
So you're saying my second equation should be .5m^2+.5v^2=.5m2^2+.5v2^2?
No, it should be:
0.5ma*(va1)^2+0.5mb*(vb1)^2 = 0.5ma*(va2)^2+ 0.5mb*(vb2)^2

(Of course, you call va2 = X and vb2 = Y.)

If you substitute the expression you found from the first equation, which gave you Y in terms of X, into this equation, then you can solve for X. And then use it to get Y.
 
after we have this equation: let's call in 1 ->(17*32)+(35*12)=(17*x)+(35*y)
we also use (V1i-V2i)=-(v1f-v2f) => 32-12+v1f=v2f -(2)

sub equation 2 into 1 and we get the final answer
am i correct?
pls advice :D
 
phynoob said:
we also use (V1i-V2i)=-(v1f-v2f) => 32-12+v1f=v2f -(2)
Yes, you can definitely use that relationship. The relative velocity is reversed in an elastic collision. Note that this relationship is derived from momentum and energy conservation. It's certainly quicker to use this formula, since much of the work has been done for you.

But some courses do not cover it, so I didn't want to bring it up. (You can always derive it for yourself, of course. :wink:)
 
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