Question

1.) A flat, circular disk of uniform thickness has a radius of 5.00 cm. A hole is drilled in the disk that is 2.50 cm in radius. The hole is tangent to one side of the disk. Where is the x-component of the centre of mass of the disk now that the hole has been drilled?

2.a) A 0.100 kg ball collides elastically with a 0.300 kg ball that is at rest. The 0.100 kg ball was traveling in the positive x-direction at 6.30 m/s before the collision. What is the velocity of the 0.300 kg ball after the collision? If the velocity is in the negative x-direction, enter a negative value.

2.b) What is the velocity of the 0.100 kg ball after the collision? If it is in the negative direction, enter a negative value.

Answer 1

Explanations & Calculations

1. )

• Since all the sections are in the same plane there is no need of assessing the y component of the new center of mass which is obvious as it is not even questioned.
• The x component by the way should lie on the line passing through both the centers of the disc & the hole.
• Before drilling the hole, the center of mass was in the center of the disc & it should shift to the side opposite to where the hole is because more mass is now concentrated on that side.
• With all these in mind, we can sketch something like,

• Imagine the masses of the components are according to the picture.
• Then balancing torques about the center of the disc, we are able to get the result

$\qquad\qquad \begin{aligned} \small Mgx&=\small mg\times2.5\\ \small x&=\small \frac{2.5m}{M} \end{aligned}$

• Since there is no evidence for the masses of each component the answer can be given as the new center of mass lies on the opposite side of the hole & the center of the disc on the line that passes through the centers.

2)

• Consider the velocity of the 0.1 kg ball before the collision to be (already given) $\small 6.30\to$ and those of the 0.1kg & 0.3kg balls after the collision to be $\small u\to \,and\,v\to$ respectively.
• Then applying linear momentum conservation,

$\qquad\qquad \begin{aligned} \small 0.1\times6.30+0.3\times0&=\small 0.1u+0.3v\\ \small 6.3&=\small u+3v\cdots(1) \end{aligned}$

• Since the collision is elastic, the kinetic energies of the balls are also conserved. Then,

$\qquad\qquad \begin{aligned} \small \frac{0.1\times(6.30)^2}{2}&=\small \frac{0.1u^2}{2}+\frac{0.3v^2}{2}\\ \small 39.69&=\small u^2+3v^2\cdots(2) \end{aligned}$

• Solving for $\small v$ yeilds,

$\qquad\qquad \begin{aligned} \small v&=\small \bold{+3.15\, ms^{-1}} \end{aligned}$ (along positive x direction)

• Then solving for $\small u$ in (1),

$\qquad\qquad \begin{aligned} \small u&=\small \bold{-3.15\,ms^{-1}} \end{aligned}$ (along negative x direction)