How Does a Coefficient of Restitution of 1/3 Affect Post-Collision Speeds?

AI Thread Summary
In a collision involving two equal mass balls moving at 3 m/s, the coefficient of restitution is given as 1/3. The correct application of the restitution formula reveals that the final velocities can be derived from the equations of conservation of momentum and the coefficient of restitution. After resolving the equations, it is determined that both balls will have a speed of 1 m/s post-collision. The confusion arose from incorrect input of velocities into the restitution equation, highlighting the importance of correctly identifying initial velocities. Ultimately, the solution confirms that the final speed for each ball after impact is indeed 1 m/s.
arpitm08
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Coefficient of Restitution? PLEASE HELP!

Homework Statement



Two balls of equal mass moving with the speed of 3 m/s, collide head on. Find the speed of each after impact if the coefficient of restitution is 1/3.

Homework Equations



e=(v2-v1)/(u1-u2)
v2= velocity after impact of object 2
v1= velocity after impact of object 1
u2= velocity before impact of object 2
u1= velocity before impact of object 1

The Attempt at a Solution



I put the numbers into the equation and it yield 0 over something.
The answer is 1 m/s for both ball, but with that into the equation, it would yield 0/0.
My book doesn't say anything on how to do this problem. Maybe it is as simple as multiplying the velocities by 1/3 to get the answer.
Could someone please help me out?
 
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Since the balls approach each other, one has a velocity of +3 m/s while the other has a velocity of -3 m/s.
 
yes that is true, but i just copied the problem exactly from the book.
 
arpitm08 said:
yes that is true, but i just copied the problem exactly from the book.
There's nothing wrong with the problem statement, just with your attempted solution.
I put the numbers into the equation and it yield 0 over something.
That just means you input the wrong numbers. Realize that u and v are velocities, not just speeds.
 


Considering
u1 = 3 m/s
u2 = -3 m/s
e = 1/3
e = (v2-v1)/(u1-u2)
→ v2-v1 = 2 m/s
u1+u2=v1+v2
...(conservation of linear momentum)
→ v1+v2 = 0
→ v1 = -v2
Therefore, the speed of each ball after impact is 1 m/s.
 
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