# Another Circular motion Question

• steejk
In summary, after being projected from the top with a speed of u, bead A collides with bead B. The coefficient of restitution is 0.5, and from the conservation of momentum, it is determined that v1 is the final velocity of bead A.
steejk
A smooth wire forms a circular hoop of radius 1m. It is fixed in a vertical plane. Two beads, A and B, of masses m and 2m respectively, are threaded onto the wire. The coefficient of restitution between the beads is 0.5. Bead B rests at the bottom of the hoop. Bead A is projected from the topmost point with speed u, and subsequently collides with B.

Q. Show that Bead A is brought to rest by the collision.

My working so far:

Using the energy change of GPE --> KE

GPE = 2mg = 0.5mv^2

Increase in velocity is therefore 2sqrt(g)

At the point A collides with B its velocity is therefore u + 2sqrt(g)

Using conservation of momentum,

u + 2sqrt(g) = 2x + y (where x is the speed of B and y is speed of A after collision)

Using Newton's law of restituiton:

0.5 = (2x-y)/(u + 2sqrt(g))

And this is where I get stuck :( ...

GPE = 2mg = 0.5mv^2

Increase in velocity is therefore 2sqrt(g)

Redo the algebra 2$\sqrt{g}$ is not the increase in velocity.

u + 2sqrt(g) = 2x + y (where x is the speed of B and y is speed of A after collision)

Remember, momentum is mass * velocity. So, on the left should be m*(u + increase in velocity). The right side also is missing the masses of the beads.

Lastly, the equation for the coefficient of restitution deals with initial and final velocities, not, momentum.

wbandersonjr said:
Redo the algebra 2$\sqrt{g}$ is not the increase in velocity.

Any clues on where I'm going wrong..?

wbandersonjr said:
Remember, momentum is mass * velocity. So, on the left should be m*(u + increase in velocity). The right side also is missing the masses of the beads.

Lastly, the equation for the coefficient of restitution deals with initial and final velocities, not, momentum.

The masses cancel in the momentum equation I think. Would the initial velocity then not be u + increase. I did the final velocity wrong, is it just x - y?

You should calculate the total kinetic energy of bead A immediately before impact in order to determine its velocity. The change in velocity is not 2sqrt(g).

I suggest that you create another symbol to represent the impact velocity of bead A, say v, and leave it that way until the end when you may (or may not ) need it expanded into its full form.

So, calling the initial speed of bead A v, the conservation of momentum gives you:

m*v = m*v1 + 2m*v2

where v1 is the speed of bead A after collision, and v2 is the speed of bead B. Clearly you can cancel the m's from that expression. Now write your Restitution Law in terms of v1 and v2 and see how things progress.

gneill said:
You should calculate the total kinetic energy of bead A immediately before impact in order to determine its velocity. The change in velocity is not 2sqrt(g).

I suggest that you create another symbol to represent the impact velocity of bead A, say v, and leave it that way until the end when you may (or may not ) need it expanded into its full form.

So, calling the initial speed of bead A v, the conservation of momentum gives you:

m*v = m*v1 + 2m*v2

where v1 is the speed of bead A after collision, and v2 is the speed of bead B. Clearly you can cancel the m's from that expression. Now write your Restitution Law in terms of v1 and v2 and see how things progress.

Okay, so from the total KE at bottom v (speed of A before collision)

v^2 = u^2 + 4g

And from restitution,

0.5v = v2 - v1

Do I subst. v into this then?

use the equation gneill gave:

m*v = m*v1 + 2m*v2

And yes, the m's cancel, I didnt see that you did that, sorry. Anyway use that equation and
0.5v = v2 - v1

Solve for v1, the final velocity of bead A. Pretend v is known and just run through the algebra, you can solve one equation for v2 and substitute it into the other equation.

wbandersonjr said:
use the equation gneill gave:

And yes, the m's cancel, I didnt see that you did that, sorry. Anyway use that equation and

Solve for v1, the final velocity of bead A. Pretend v is known and just run through the algebra, you can solve one equation for v2 and substitute it into the other equation.

Got it :) Thanks wbandersonjr and gneill

## 1. What is circular motion?

Circular motion is the movement of an object in a circular path around a fixed point. This type of motion is often seen in objects that are rotating or moving in a circular orbit.

## 2. What causes circular motion?

Circular motion is caused by a centripetal force, which is a force that acts towards the center of the circular path and keeps the object moving in a circular motion. This force can come from various sources, such as gravity, tension, or friction.

## 3. How is circular motion different from linear motion?

Circular motion involves an object moving around a curved path, while linear motion involves an object moving in a straight line. In circular motion, the direction of the object's velocity is constantly changing, while in linear motion, the velocity remains constant.

## 4. What are some real-life examples of circular motion?

Examples of circular motion in everyday life include the rotation of the Earth around the sun, the orbit of the moon around the Earth, the motion of a spinning top, and the movement of a car around a roundabout.

## 5. How is circular motion related to other types of motion?

Circular motion is a type of rotational motion, which is the movement of an object around its own axis. It is also related to periodic motion, which is any motion that repeats itself after a certain amount of time. Circular motion can also be broken down into simpler forms of motion, such as linear motion and simple harmonic motion.

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