The Student Room Group

m2 question

A particle A of mass 2m is moving with speed 2u on a smooth horizontal table. The particle collides directly with a particle B of mass 4m moving with speed u in the same direction as A. The coefficient of restitution between A and B is 0.5.

(a) Show that the speed of B after the collision is 2/3u

(b) Find the speed of A after the collision. which im sure is just u

Subsequently B collides directly with a particle C of mass m which is at rest on the table. The coefficient of restitution between B and C is e. Given that there are no further collisions,

(c) find the range of possible values for e. (im not quite sure how to do this one, any help please?)
Reply 1
anyone please?
Reply 2
anyone?
Only maths exams, C2, C3 and M2 left; so i'll be in the maths forums often if you need my help.

A particle A of mass 2m is moving with speed 2u on a smooth horizontal table. The particle collides directly with a particle B of mass 4m moving with speed u in the same direction as A. The coefficient of restitution between A and B is 0.5.

(a) Show that the speed of B after the collision is 2/3u

mA = 2m
mB = 4m
uA = 2u
uB = u
vA = vA
vB = vB

Conservation of momentum
2m x 2u + 4m x u = 2m x vA + 4m x vB
4mu + 4mu = 2mvA + 4mvB
8mu = 2mvA + 4mvB
8u = 2vA + 4vB (1)

Coefficient of restitution
e = (vB - vA)/(2u - u) = (vB - vA)/u
vB - vA = eu
vA = vB - eu (2)

(2) into (1)
8u = 2(vB - eu) + 4vB
8u = 2vB - 2eu + 4vB
6vB = 8u + 2eu
6vB = 8u + 2(½)u
6vB = 9u
vB = 9u/6 = 3u/2

I think this is right.

(b) Find the speed of A after the collision.

vA = vB - eu
vA = vB - ½u
vA = 3u/2 - ½u
vA = u

Subsequently B collides directly with a particle C of mass m which is at rest on the table. The coefficient of restitution between B and C is e. Given that there are no further collisions,

(c) find the range of possible values for e.

mB = 4m
uB = 3u/2
vB = vB

mC = m
uC = 0
vC = vC

e = e

If there are no further collisions after this collision vB > u because otherwise A would catch up with B and collide again.

Conservation of momentum
4m x 3u/2 + m x 0 = 4m x vB + m x vC
6mu = 4mvB + mvC
6u = 4vB + vC
vC = 6u - 4vB (3)

Coefficient of restitution
e = (vC - vB)/3u/2
e = 2(vC - vB)/3u
3eu/2 = vC - vB
3eu/2 = 6u - 4vB - vB
5vB = 6u - 3eu/2
vB = 6u/5 - 3eu/10

vB > u
=> 6u/5 - 3eu/10 > u
=> 6u/5 - u > 3eu/10
=> u/5 > 3eu/10
=> 2u > 3eu
=> 3eu < 2u
=> 3e < 2
=> e < 2/3
Reply 4
thanks, rep coming your way but it says "you must spread rep points around before you add to widowmaker's" ....

where i went wrong was doing vB<vC instead of using u, and then it all got confusing, but thanks i get it now!