
Arturo O. answered 08/09/17
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I will set this up for you and you can work the math.
(a) Use energy conservation.
mAgL(1 - cosθ) = mvA2/2
vA = √[2gL(1 - cosθ)] = √[2(9.8)(0.30)(1 - cos60°)] ≅ 1.71 m/s
(b) Use conservation of linear momentum and the definition of elastic collision.
vAi = result for vA from part (a).
mAvAi = mAvAf + mBvB ⇒ 0.040(1.71) = 0.040vAf + 0.060vB
0.0684 = 0.040vAf + 0.060vB
mAvAi2/2 = mAvAf2/2 + mBvB2/2 ⇒ 0.040(1.71)2 = 0.040vAf2 + 0.060vB2 ⇒
0.117 = 0.040vAf2 + 0.060vB2
Solve the following 2 equations in 2 unknowns for vAf and vB:
0.0684 = 0.040vAf + 0.060vB
0.117 = 0.040vAf2 + 0.060vB2
This is algebra now. You should be able to get vAf and vB from here, with vAf negative (moving to left) and vB positive (moving to right).
(c) Use energy conservation. Using the results from part (b),
mghAmax = mvA2/2 ⇒ hAmax = vA2/(2g)
Similarly,
hBmax = vB2/(2g)
You can finish from here.