Elastic Collision of a 90kg ball

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SUMMARY

The discussion centers on the analysis of an elastic collision involving a 90g ball moving at 100 cm/s colliding with a stationary 10g ball. The solution for part (a) indicates that if the balls stick together, their combined velocity post-collision is 90 cm/s. For part (b), the correct approach involves applying both conservation of momentum and conservation of kinetic energy to determine the final velocities of the two balls, which results in the 10g ball moving at 80 cm/s and the 90g ball moving at 20 cm/s after the elastic collision.

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Lyphta
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Homework Statement


A 90-g ball moving at 100 cm/s collides head-on with a stationary 10-g ball. Determine the speed of each after impact if (a) they stick together, (b) the collision is perfectly elastic.

Homework Equations


p=mv

The Attempt at a Solution


I already solved A, but I don't know how to do B. So far, I've gotten:

p=mv
= 90(100)
= 9000 g.cm/s

p=mv
= 10 (0)
= 0

so...
p1+p2=p3
m1 v1 + m2 v2 = (m1 +m2)v
90(100) + 0 = 100v
v= 90 cm/s

I don't understand how the answer can be 80 cm/s when they perfectly elastic. Shouldn't it be 90 cm/s?
 
Last edited:
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p1+p2=p3
m1 v1 + m2 v2 = (m1 +m2)v
90(100) + 0 = 100v
v= 90 cm/s

This answer is for (a). In the part (b) after impact both bodies will have different velocities. So wrigth down two equations. One for coservation of linear momentum and another for conservation of energy. Then solve for the new velocities of the two bodies.
 
Lyphta said:

The Attempt at a Solution


I already solved A, but I don't know how to do B. So far, I've gotten:

One method has already been given. You can also use the fact that the relative velocity before collision is equal to the relative velocity after collision, because the collision is fully elastic.
 

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