20 3.[1pt] Two identical steel balls, each of mass 2.8 kg, are suspended from st

Question

20 3.[1pt] Two identical steel balls, each of mass 2.8 kg, are suspended from strings of length 29 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle theta - 63° with the vertical and let it go It collides elastically with the other ball How high will the other ball rise above its starting point? Answer 4.[1pt] Suppose that instead of steel balls we use putty balls. They will collide inelastically and remain stuck together after the collision. How high will the balls rise after the collision? Answer

Explanation / Answer

According to the given problem,

3.)Since the collision is elastic then KE is not lost by the system. This KE is equal to PEg of the steel ball which was pulled so its string make an angle 53 deg with the vertical.
PEg = mgh. Solving for h.
h = length of string - cos63oxlength of string
h = 0.29m - 0.29cos63o
h = 0.16 m
PEg = 2.8kg x 9.8 m/s2 x 0.12 m
PEg = 4.35 J = KE of swinging ball when it strikes the stationary steel ball.

Now, this KE = 4.35 J will disappear until it reappers as PEg when the hit ball swings upward.
It is clear that the height it will go up is the same height as before: 0.16 m.

4.)KE of swinging putty ball before it hits the stationary putty ball = 4.35 J (see solution above.)
KE = 1/2(m)v2
4.35 = (1/2)(2.8kg)v2
v = 1.76 m/s

m1v1 + m2v2 = (m1+m2)v
2.8(1.76) + 2.8(0) = (2.8+ 2.8)v
v = 0.88 m/s
KE of 2 puttballs = PEg of two putty balls
(1/2)(5.6)(0.88)2 = (5.6)(9.81)h
h = 0.04 m or 4cm .

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