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?)
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)
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)