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Coefficient of restitution(M2)

I really stuck on this. In the equation is the sign always -e, or does it change because of the direction in which the particles are moving. Would be really greatful for an answer and explanation, Thanks!!
Reply 1
Avatar for phys981
phys981
10
If i understand this, the coefficient of restitution is the ratio of the rebound velocity to the approach velocity?

In which case it will always be negative just because these two directions are opposite.

I suppose if its given as positive then you´re only concerned with the speeds, not the directions, of rebound and approach.
e has no direction. its just a ratio really. e= (speed or separation)/ (speed of approach). And notice its speed not vel.

it doesnt matter which way the particles are travelling eg if you have A 1m/s to right B 5m/s to right, the speed of approach= 4
if they separate at 1m/s and 2m/s in opp directions, speed of separation is 3m/s. its the overall difference in speed.

then e= 3/4 regardless of which way they are travelling.
e is between 0 and 1. 0 is completely inelastic and the particles stick. 1 is perfectly elastic collisions.
Reply 4
Avatar for phys981
phys981
10
Aha, so then the -ve that crystal_v has is there to calculate velocity

ie V1 = -eV0 or e = V1/V0

The e IS just a number so you add a minus sign when you calculate veolocity?
the question will tell you or imply the direction of travel, so you'll know when its negative. often you have to apply conservation of momentum so direction matters. e is just a ratio but cannot be negative.

so if e=1/2 against a wall, and its approaching at 5ms/1, the rebound speed will be 2.5m/s. depending on your sign convention, its velocity is now negative velocity.
but sometimes objects may collide but continue to move in the same direction so you cant assume its negative.
Reply 6
Avatar for spoonboy
spoonboy
Always take right to be the positive direction.
If a particle moves to the left then its velocity must be is negative.
If a particle moves to the right then its velocity must be is positive.

Particle A has initial velocity u1
Particle B has initial velocity u2
Particle A has final velocity v1
Particle B has final velocity v2
Initially Particle A is to the left of Particle B. They collide.
For the particles to approach, u1 must be greater than u2
For the particles to seperate, v2 must be greater than v1
u1 - u2 is the approach speed, and is always positive
v2 - v1 is the separation speed, and is also always positive

e = (seperation speed) / (approach speed)
e = (v2 - v1) / (u1 - u2)
which is always positive
Reply 7
Avatar for glasses
glasses
10
so basically in any given situation the coefficient of restitution is positive? is this because we are only looking at speeds and not velocities?

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