Question

A 6.00 kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center. The bar has mass 5.00 kg and is 4.00 m in length. At the other end of the bar sits another 6.00 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.

Answer 1

Answer:

The height of the other ball go after the collision is 2.304 m.

Explanation:

Given that,

Mass of ball = 6.00 kg

Height = 12.0 m

Mass of bar =5.00 kg

Length = 4.00 m

Suppose we need to calculate how high will the other ball go after the collision

We need to calculate the velocity of ball

Using formula of velocity

$v=\sqrt{2gh}$

$v=\sqrt{2\times9.8\times12.0}$

$v=15.33\ m/s$

We need to calculate the angular momentum

Using formula of angular momentum

$l_{before}=mvr$

Put the value into the formula

$l_{before}=6.00\times15.33\times2.0$

$l_{before}=183.96\ kgm^2/s$

We need to calculate the angular momentum

Using formula of angular momentum

$l_{after}=I_{t}\omega$

$l_{after}=(\dfrac{ml^2}{12}+m_{1}r^2+m_{2}r^2)\omega$

Put the value into the formula

$l_{after}=(\dfrac{5\times4.00^2}{12}+6.00\times2.0^2+6.00\times2.0^2)\omega$

$183.96=54.66\omega$

$\omega=\dfrac{183.96}{54.66}$

$\omega=3.36\ rad/sec$

After collision the ball leaves with velocity

We need to calculate the velocity after collision

Using formula of the velocity

$v= r\omega$

$v=2.0\times3.36$

$v=6.72\ m/s$

We need to calculate the height

Using formula of height

$h=\dfrac{v^2}{2g}$

Put the value into the formula

$h=\dfrac{(6.72)^2}{2\times9.8}$

$h=2.304\ m$

Hence, The height of the other ball go after the collision is 2.304 m.