The maximum height an object on a wire moves after a collision

[imatreyu]

Homework Statement

A. 04 kg bead slides on a curved frictionless wire, starting from rest at point A. At point B, the bead collides elastically with a 0.6-kg ball at rest. Find the height that the ball moves up the wire.

The answer is .96 m.

Homework Equations

m1v1i + m2v2i=m1v1f + m2v2f

The Attempt at a Solution

PE lost by A = KE gained by A by the time it collides with B-->

mgh=1/2mvf^2
0.4 (9.8)(1.5) = 1/2 (0.4)vf^2
vf= square root ((2)(0.4)(9.8)(1.5)/.4)
vf= 5.422 m/sSo I use the equation for elastic collisions. . .

0+ 0.6 (5.422)= 0.4(5.422)+ 0.6 v2f

v2f= 1.8074 m/s

Then I plug in: mgh = 1/2mv^2

.6 (9.8) h = (1.8072^2) .5 .6
h= .16 m

THIS IS WRONG H IS SUPPOSED TO BE .96 m.

I don't know what I'm doing wrong!

 

[rl.bhat]

0+ 0.6 (5.422)= 0.4(5.422)+ 0.6 v2f

This equation is wrong. It should be

$$m_1v_{1i}+ m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$

Similarly write down the equation for the conservation of energy. Using these two equations solve for final velocities of both the masses.

 

[imatreyu]

Hello! Thank you for your response.

Using a simplified equation derived from the conservation of energy (Serway & Faughn, 1992):

v1i + v1f = v2f +v2i----> v1f= v2f + v2i - v1i
m1v1i + m2v2i = m1(v2f + v2i - v1i) + m2v2f

I don't know what to plug in. I plug things in, but I am getting an incorrect answer. The answer should be -1.084. I am not getting this answer. Once I find the correct v2f I should be able to perform the last step correctly. . .

 

[rl.bhat]

Using the conservation of momentum and energy, you will get

$$v_{1f} = \frac{v_{1i}(m_1 - m_2)}{(m_1 + m_2)}$$

$$v_{2f} = \frac{2v_{1i}m_1}{(m_1 + m_2)}$$

Now proceed.