Work - energy theorem and conservation of momentum question

[hermy]

Homework Statement

There is a hemispherical track of radius R = 4m. A block B is lying at rest at the bottom of the track. A block A is pushed along the horizontal portion of the track with speed u. Collision between A and B is perfectly elastic. Find u such that B reaches P (the end of hemispherical portion of track). (See figure for clarity)

Given: mass of A = 2 kg
mass of B = 4 kg

track is frictionless

The Attempt at a Solution

velocity of B after collision should be $$\sqrt{}$$2gRlet velocity of A after collision be v

u = v + 2$$\sqrt{}$$2gR

$$\sqrt{}$$2gR - v = u

------------------------------------------ adding
3$$\sqrt{}$$2gR = 2u

u = 12 m/s (approx)
by conserving momentum and condition for elastic collision, u = 12 m/s (approx)

is the solution correct? the answer is supposed to be 10 m/s.

 

[Doc Al]

Where's your solution?

 

[Doc Al]

velocity of B after collision should be $$\sqrt{}$$2gR


let velocity of A after collision be v

u = v + 2$$\sqrt{}$$2gR

This looks good.

$$\sqrt{}$$2gR - v = u

Where did you get this?

 

[hermy]

Thanks doc, for showing interest in this question.

$$\sqrt{}$$2gR - v = u

Where did you get this?

for elastic collision, coefficient of restitution is 1.

so,
velocity of separation = (-1) velocity of approach

$$\sqrt{}$$2gR - v = -( 0 - u)
= u

 

[Doc Al]

OK, I see what you did. Makes sense.

The problem is that you found the pre-collision speed of A at the bottom of the track, but they want the speed while it's still on the horizontal portion. Redo it with that in mind and you'll get the right answer.

 

[hermy]

Thanks, i got it. I got to avoid such mistakes, really...