Question

A 200 g rubber ball is attached to a 1.0 m long string and released from an angle (theta). It swings down and at the very bottom has a perfectly elastic collision with a 1.0 kg block. The block is resting on a frictionless surface and is connected to a 20 cm long spring with spring constant 2000 N/m. After the collision, the spring compresses a maximum distance of 3.0 cm. From what angle was the ball released?

Answer #1

We know that the height the ball drops from is given as h = L*(1 - cos), so

Ep = mgh = mgL(1-cos) = 0.2kg *
9.8m/s^{2} * 1m * (1 - cos) = 1.96J * (1 -
cos)

This is converted into Ek at the bottom of the arc:

Ek = 1.96J * (1 - cos) = ½mv² = 0.1kg
* v^{2}

v² = 19.6m^{2}/s^{2} * (1 - cos)

v = 4.43m/s * sqrt(1 - cos)

Let K = sqrt(1 - cos).

B. initial momentum = mv = 0.2kg * 4.43m/s * K = 0.89kg·m/s * K

final momentum = 0.89kg·m/s * K = 0.2kg * u + 1kg * v

For a perfectly elastic collision, we know that

relative velocity of approach = relative velocity of separation, so

4.43m/s * K = v - u, or

v = u + 4.43m/s * K

Plug this into the momentum eqn:

0.89kg·m/s * K = 0.2kg * u + 1kg * (u + 4.43m/s * K)

0.89kg·m/s * K - 4.43kg·m/s * K = 1.2kg * u

u = -3.54kg·m/s * K / 1.2kg = -2.95m/s * K ---- velocity of ball

v = u + 4.43m/s * K = 1.48m/s * K -----------velocity of block

For the spring, Ep =
0.5*kx^{2} =0.5 * 2000N/m * (0.03m)^{2} = 0.9
J

So this must equal the initial Ek of the block:

0.9 J = 0.5*mv^{2} =0.5 * 1kg * (1.48m/s *
K)^{2} = 1.1J * K^{2} = 1.1J * (1 - cos)

1 - cos = 0.81

cos = 0.19

=
79.04^{0}

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