Conservation of Energy and ball problem

[cdbowman42]

1.Ball 1 with an initial speed of 14 m/s has a perfectly elastic collision with Ball 2 that is initially at rest. Afterward, the speed of Ball 2 is 21 m/s. What will be the speed of Ball 2 if the mass of Ball 1 is doubled.



2. conservation of momentum: m₁(v_1i)=m₁(v_1f)+m₂(v_2f)

conservation of energy:.5m₁(v_1i)²=.5m₁(v_1f)²+.5m₂(v_2f)²

v_1f=v_1i(m₁-m₂)/(m₁+m₂)

v_2f=v_1i(2m₁)/(m₁+m₂)



3. I'm not sure where to even begin because it seems I would need to know the mass of either ball to find the solution.

 

[ldamewood]

The problem provides the initial speed of ball 1 and the final speed of ball 2, you should be able to use those numbers to find the ratio of the ball's masses m2/m1. Once you find this quantity, you have enough information to determine the final speed of ball 2 if m1 doubles.

Hint: Solve for a=m2/m1 by creating ratios:
$$b = \frac{v_{2f}}{v_{1i}} = \frac{2m_1}{m_1+m_2} = \frac{2}{1+a}$$
where I'm using a and b to replace the ratio quantities. Now you have a very simple algebraic equation.

 

[iRaid]

p1+p2=p1'+p2'
p2 = m2v2 = 0 (v=0)
Elastic means that the balls don't stick so p1' = 0 (v=0, again).

so you're left with p1=p2'
m1v1 = m2'v2'

the m's are proportional so doubling m1 will double m2 which means ? (you answer)

 

[ldamewood]

p1+p2=p1'+p2'
p2 = m2v2 = 0 (v=0)
Elastic means that the balls don't stick so p1' = 0 (v=0, again).

so you're left with p1=p2'
m1v1 = m2'v2'

Elastic does mean that the balls don't stick, however your analysis presupposes that the masses are identical. If they are not identical, then p1' is not zero. In fact, just by looking at the speeds given, you should be able to determine that m1 is a bit more massive than m2 (before you double the mass of m1).