What Is the Wagon's Speed After an Elastic Collision?

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SUMMARY

The discussion centers on calculating the speed of a wagon after an elastic collision involving a 150 kg cart and a 420 kg wagon. The cart, initially moving at 13 m/s east, rebounds westward at 3 m/s after the collision. The correct approach to find the wagon's speed is through the conservation of momentum, yielding a final speed of 9.0 m/s, while the provided answer of 11 m/s is incorrect. The discussion emphasizes that conservation of kinetic energy applies only when the coefficient of restitution is equal to 1, which is not specified in this scenario.

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Homework Statement



A 150 kg cart moving at 13 m/s east collided with a 420 kg wagon moving at 5.0 m/s east.
The cart rebounded westward with a speed of 3.0 m/s . What was the speed of the wagon after the collision? Observe standard Cartesian coordinates.

Homework Equations



KE1+KE2=KE1'+KE2'
conservation of momentum
conservation of kinetic energy in elastic collisions

The Attempt at a Solution


(0.5)(150)(13)2+(0.5)(420)(5)2=(0.5)(150)(-3)2+(0.5)(420)v2
v=9.0m/s

Answer as given is 11m/s
 
Last edited:
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the problem itself doesn't say that the collision is elastic. so just use conservation of momentum here.
 
canicon25 said:

Homework Statement



A 150 kg cart moving at 13 m/s east collided with a 420 kg wagon moving at 5.0 m/s east.
The cart rebounded westward with a speed of 3.0 m/s . What was the speed of the wagon after the collision? Observe standard Cartesian coordinates.

Homework Equations



KE1+KE2=KE1'+KE2'
conservation of momentum
conservation of kinetic energy in elastic collisions


The Attempt at a Solution


(0.5)(150)(13)2+(0.5)(420)(5)2=(0.5)(150)(-3)2+(0.5)(420)v2
v=9.0m/s

Answer as given is 11m/s

IssacNewton is right !
Conservation of kinetic energy is valid only when coefficient of restitution(e) is 1
here its not specified, so you must use momentum conservation.

ALSO:
if e is given and is ≠1 then you may use:
(v2 - v1) = e(u1 - u2)
 

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