Two balls, of masses mA=45g and mB=65g, are suspended

[josephyarwick]

Homework Statement

Two balls, of masses mA=45g and mB=65g, are suspended as shown in figure 9-53. The lighter ball is pulled away to a 66 degree angle with the vertical and released. (a.) What is the velocity of the lighter ball before impact? (b.) What is the velocity of each ball after the elastic collision? (c.) What will be the maximum height of each ball after the elastic collision?
mA=.045kg mB=.065 theta(0)=66 degrees L=.3m

Homework Equations

1/2mv^2+mgh=1/2mv^2+mgh
1/2mv^2+1/2mv^2=1/2mv^2+1/2mv^2

The Attempt at a Solution

I'm very lost, I originally tried to find (a.) by setting v=(sqrroot)(2g(.3)sin(66)) but I got 2.3 m/s which is supposedly incorrect. I then tried replacing sin(theta) with (1-cos(theta)) and somehow got the right answer, 1.9 m/s. For part (b.) I tried to find the velocity immediately after the collision by solving for one of the v(prime) in both the conservation of kinetic energy and the conservation of momentum equations. I then substituted the results but I my answer was .9 m/s for v(prime)A when it was supposed to be -.3 m/s. please help

 

[gneill]

What's your logic behind your attempt at (a)? It would help if you could include the referenced diagram.

 

[josephyarwick]

What's your logic behind your attempt at (a)? It would help if you could include the referenced diagram.

the diagram is simply two balls (of masses mA and mB) suspended on two separate pendulums of equal length L=.3 m with the lighter ball (ball mA) held up at 66 degrees to the left.
since 1/2mv^2=mgh. I figured that I could place L-Lcos(theta) to make the height dependent on the angle. after simplifying, I got v=(sqrroot)(2(9.81)(.3-.3cos(66))
This seems to be right answer. but when I plus it into my approach at (b) everything falls apart

 

[gneill]

Can you show the math details of your work for part b?

Note that it's often easier to work with the closing / separation velocity relationship rather than the KE conservation relationship when doing elastic collisions. It states that the speed of separation after collision is equal to the speed of closing before the collision. Or putting it in mathematical terms, if the initial velocities of two objects are v1 and v2, and their final velocities are u1 and u2, then

v1 - v2 = u2 - u1

You don't have to deal with any velocity squares that way.

 

[josephyarwick]

Or putting it in mathematical terms, if the initial velocities of two objects are v1 and v2, and their final velocities are u1 and u2, then

v1 - v2 = u2 - u1

OOOHHH, I understand now, so if I use this equation instead of the conservation of KE, which I did before, I have the equations:
m(A)v(1)+m(B)v(2)=m(A)v'(1)+m(B)v'(2) and v(1)-v(2)=v'(2)-v'(1)

then I solve for either v'(1) or v'(2) and cancel out where the velocity is zero at v(2)

v'(1)=(m(A)v(1)-m(B)v'(2))/m(A) and v'(1)=v'(2)-v(1)

then substitute and solve for v'(2)

(m(A)v(1))/m(A) - (m(B)v'(2))/m(A) = v'(2)-v(1)
(m(A)v(1))/m(A) + v(1) = v'(2) + (m(B)v'(2))/m(A)
v'(2) = (v(1) + v(1))/(1 + m(B)/m(A))
2(1.87m/s)/(1+.065/.045) = 3.74/2.44 ~ 1.5 m/s
v'(1) = 1.54 - 1.87 ~ -0.3 m/s

Thank you so much!