How would you solve these two problems? (AP Physics)?

7-27

"A steel ball of mass m is fastened to a light cord of length L and released when the cord is horizontal. At the bottom of its path, the ball strikes a hard plastic block of mass M = 4m, initially at rest on a frictionless surface. The collision is elastic.

"(a) Find the tension in the cord when the ball's height above its lowest position is 0.5L. Write your answer in terms of m and g."

From Conservation of Energy, we know v² = 2gh = 2g(0.5L) = gL.

Tension T = m(v²/L + gcosΘ)

where Θ = 60º from vertical when h = 0.5L.

Since cos60º = ½, we have

T = m(gL/L + g/2) = 1.5mg

"(b) Find the speed of the block immediately after the collision."

Just before impact, v = √(2gL). Conserve momentum:

mv = mU + 4mV

for U, V the post-collision velocities of the ball, block respectively

m cancels, leaving

v = √(2gL) = U + 4V

For an elastic, head-on collision, we know (from CoE) that

the relative velocity of approach = relative velocity of separation, or

√(2gL) = V - U, so

U = V - √(2gL) → plug this into the momentum equation

√(2gL) = V - √(2gL) + 4V

2√(2gL) = 5V

V = 0.4√(2gL)

"(c) To what height h will the ball rebound after the collision?"

U = V - √(2gL) = (0.4 - 1)√(2gL) = -0.6√(2gL)

and the ball rises to height

h = U² / 2g = (-0.6√(2gL))² / 2g = 0.36*2gL / 2g = 0.36L

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