Coefficient of restitution of a particle

[erisedk]

Homework Statement

A particle moving with initial velocity vi = (3i + 5j) m-s−1, collides with a smooth plane wall placed at some orientation to the particle’s trajectory. The resulting velocity of the particle is vf = (−2i − j) m-s−1. The coefficient of restitution for this collision is
Ans: 16/45

Homework Equations

e = velocity of separation/velocity of approach

The Attempt at a Solution

So, my first attempt was
e = [ 0 - (−2i − j) ]/ [ (3i + 5j) - 0 ] = (2i + j)/(3i +5j)
I thought of taking the speeds then, which gave me e = √5 / √34
Which is obviously not the answer.
Then I tried to google COR, and i got-- "representing the ratio of speeds after and before an impact, taken along the line of the impact"

I have a feeling that my answer is wrong because I'm not incorporating "line of impact". But I have no idea how to, because I'm not given the angle at which the particle collides with the wall.

 

[Orodruin]

There is an underlying assumption that the wall can only provide a force (and thus acceleration) perpendicular to itself (the keyword is "smooth"). You can use this to find out how the wall is tilted.

 

[erisedk]

Could you please elaborate a bit more?

 

[Orodruin]

Are you able to compute the direction of the velocity change? The force is normal to the surface, so the velocity change is going to be in the normal direction of the surface and so computing the velocity change will give you the direction of the surface normal.

 

[erisedk]

Ok thank you!
I got it.
vf - vi = (-5i - 6j) = direction of surface normal
Taking the components of velocities about line of impact, vi = (-5i - 6j) . (3i + 5j) = 45 & vf = (-5i - 6j) . ( -2i-j ) = 16
Hence, e = 16/45.

 

[Orodruin]

Yes, just one comment: What you really computed was not the velocity components in the normal direction, but the magnitude of the projections of the velocities on a normal that was not normalised. Now, for this problem it does not matter as you are computing the ratio between two components and the normalisation cancels out. However, in general, you might want to normalise the surface normal when computing the component normal to the surface.