15. [2pt] Two small identical clay balls of mass m are suspended from the same p

Question

15. [2pt]
Two small identical clay balls of mass m are suspended from the same point by strings of length L. One of them is pulled sideways and up such that its string makes an angle as shown. After being released from rest, it collides with the other ball, and they remain stuck together after the collision. How high h will the two balls rise after the collision? Enter your symbolic answer as a function of m, L, g, and theta (you'll have to type the whole word since you don't have a key on your keyboard), plus physical constants such as 1.0 or 0.5. As with all symbolic answers, do not ever include units or unit conversion factors. For example, you might enter:

Explanation / Answer

When one mass is raised by angle theta then height of that ball beocmes(H) = L - LCos(theta) = L (1-Costheta)
now this ball have some potential energy = mgH = mgL(1-Cos(theta))
this potential energy get converted into the kinetic energy when reached the bottom.
mgH = (1/2mV2)
Now conservation of momentum takes place
momentum before collision = mV
where m is mass of one ball which is at angle theta and V is its velocity when it strike the other ball.
after collision both mass stuck together therefore they will move with same velocity
= (m+m)v = 2mv
now equating both momentum
2mv =mV
v = V/2 so final velocity is half of initial
now kinetic energy after collision = (1/2)(2m)(V/2)2 = (1/2)((1/2)mV2)
And we know that mgH = (1/2mV2)
so final kinetic energy = mgH/2
final potential energy = (2mgh)
equating final potetntial and kinetic
2mgh = mgH/2
h = H/4
h =(L/4)(1-Cos(theta)

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