How High Do Colliding Masses Reach in a Frictionless Bowl?

[joemama69]

Homework Statement

note the picture

A small mass m1 slides in a completely frictionless spherical bowl. m1 starts at rest at height h = ½ R above the bottom of the bowl. When it reaches the bottom of the bowl it strikes a mass m2, where m2 = 3m1, in a completely elastic collision.

a)find height of mass 2 after colision

b)find height of mass 1 after colision

Homework Equations

The Attempt at a Solution

m¹g(.5R) = .5m₁v_1o²...v_1o = (gR)^.5

m₁v<sub>1o[/SUB = m1v1 + m2v2...m1v1o[/SUB = m1v1 + 3m1v2...(gR)^.5 = v1 + 3v2...
v1 = (gR)^.5 - 3v2
v2 = ((gR)^.5-v1)/3

a) .5m2v2² = m2gh... .5(((gR)^.5-v1)/3)² = gh...hm_2 = (((gR)^.5-v1)/3)² /2g

b).5m1v2² = m1gh... .5((gR)²-3v2)² = gh...h = ((gR)^.5-3v2)²/2g)</sub>

 

[Redbelly98]

The Attempt at a Solution

m¹g(.5R) = .5m₁v_1o²...v_1o = (gR)^.5

m₁v_1o = m₁v₁ + m₂v₂...m₁v_1o = m₁v₁ + 3m₁v₂...(gR)^.5 = v₁ + 3v₂...

Okay so far. At this point you can use conservation of kinetic energy for the elastic collision, and get a second equation relating v1 and v2. From there, you can express v1 and v2 in terms of g and R.

v₁ = (gR)^.5 - 3v₂
v₂ = ((gR)^.5-v₁)/3

a) .5m₂v₂² = m₂gh... .5(((gR)^.5-v₁)/3)² = gh...h_{m_2} = (((gR)^.5-v₁)/3)² /2g

b).5m₁v₂² = m₁gh... .5((gR)²-3v₂)² = gh...h = ((gR)^.5-3v₂)²/2g)