ellastic collision w/rebound


by canicon25
Tags: collision, ellastic, w or rebound
canicon25
canicon25 is offline
#1
Jan2-12, 07:47 PM
P: 25
1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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


3. 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
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issacnewton
issacnewton is offline
#2
Jan2-12, 09:52 PM
P: 607
the problem itself doesn't say that the collision is elastic. so just use conservation of momentum here.
cupid.callin
cupid.callin is offline
#3
Jan2-12, 10:29 PM
P: 1,135
Quote Quote by canicon25 View Post
1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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


3. 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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