Question

In the figure below (Figure 1), the upper ball is released from rest, collides with the stationary lower ball, and sticks to it. The strings are both 50.0 cm long. The upper ball has a mass of 2.20 kg and it is initially 10.0 cm higher than the lower ball, which has a mass of 2.70 kg. Find the frequency of the motion after the collision. Find the maximum angular displacement of the motion after the collision.

Answer 1

(a) The frequency of the motion after the collision is 0.71 Hz.

(b) The maximum angular displacement of the motion after the collision is 16.3⁰.

Speed of the 2.2 kg ball when it collides with 2.7 kg ball

The speed of the 2.2 kg ball which was initially 10 cm higher that 2.7 kg ball is calculated as follows;

K.E = P.E

¹/₂mv² = mgh

v² = 2gh

v = √2gh

v = √(2 x 9.8 x 0.1)

v = 1.4 m/s

Final speed of both balls after collision

The final speed of both balls after the collision is determined from the principle of conservation of linear momentum.

Pi = Pf

m₁v₁ + m₂v₂ = vf(m₁ + m₂)

2.2(1.4) + 2.7(0) = vf(2.2 + 2.7)

3.08 = 4.9vf

vf = 3.08/4.9

vf = 0.63 m/s

Maximum displacement of the balls after the collision

P.E = K.E

$mgh_f = \frac{1}{2} mv_f^2\\\\h_f = \frac{v_f^2}{2g} \\\\h_f = \frac{(0.63)^2}{2(9.8)} \\\\h_f = 0.02 \ m$

Maximum angular displacement

The maximum angular displacement of the balls after the collision is calculated as follows;

$cos \theta = \frac{L - h_f}{L} \\\\cos\theta = \frac{0.5 - 0.02}{0.5} \\\\cos\theta = 0.96\\\\\theta = cos^{-1}(0.96)\\\\\theta = 16.3 \ ^0$

Frequency of the motion

$f = \frac{1}{2\pi} \sqrt{\frac{g}{L} } \\\\f = \frac{1}{2\pi } \sqrt{\frac{9.8}{0.5} } \\\\f = 0.71 \ Hz$

Learn more about maximum angular displacement here: brainly.com/question/13665036