Question

g ou drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 kg and is 4.2 m in length. At the other end of the bar sits another 3.6-kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision. Assume that the bar is horizontal when the dropped ball hits it. How high (in meters) will the other ball go after the collision

Answer 1

Answer:

0.4112 m

Explanation:

The mass of the 1st ball = 3.6 kg

The height of the 1st ball =3.5 m

The mass of the 2nd ball = 3.6 kg

Mass of the bar M = 9.9 kg

Length of the bar L = 4.2 m

The velocity of the ball when it dropped from the height is calculated by using the formula:

$\dfrac{1}{2}mv_1^2 = mgh_1 \\ \\ v = \sqrt{2gh_1} \\ \\ v =\sqrt{2\times9.8 \times 3.5} \\ \\ v = 8.283 \ m/s$

Provided that the bar is pivoted at the center and the ball is placed at the two ends, the moment of inertia for the bar is:

$I = \dfrac{1}{12}ML^2 + m_1 (\dfrac{L}{2})^2 + m_2(\dfrac{L}{2})^2 \\ \\ =\dfrac{1}{12}(9.9kg)(4.2m)^2 + [3.6 kg+3.6kg](\dfrac{4.2}{2 \ m})^2 \\ \\ = 46.305 \ kg.m^2$

The angular momentum of the system due to the ball can be determined by using the formula:

L = mvr

L = (3.6 kg) (8.283 m/s) (2.1 m)

L = 62.61948 kg. m²

Now, Using the law of conservation:

$L_i = L_f \\ \\ 62.61948 \ kg.m^2/s = I \omega$

$\omega = \dfrac{62.6198 \ kg.m^2/s}{46.305 \ kg.m^2}$

$\omega =1.352 \ rad/s$

The linear angular velocity is deduced to be:

$v = r \omega \\ \\ v = (2.1 \ m) ( 1.352 \ rad/s) = 2.839 \ m/s$

∴

the height raised by the second ball is:

$h_2 = \dfrac{v^2}{2g} \\ \\ h_2 = \dfrac{(2.839)^2}{2(9.8 \ m/s^2)} \\ \\ h_2 =0.4112 \ m$