HINTS FOR PROBLEM SET #13

Problem Set #13 – Problem 1

Part a)

If there is no friction during the collision, then there are no tangential forces acting on the balls B and C during the collision. Therefore, momentum is conserved for each ball in the tangential direction.  Hence, the velocities of B and C have to be along the line of impact after the collision.

Furthermore, we can use an argument of symmetry to conclude that the velocities of B and C after the collision must be equal in magnitude. 

Also, by an argument of symmetry, the velocity of A after the collision must be along the x direction only.

Momentum is conserved for the system of the 3 balls because there are no external forces in the plane of motion during the collision.  So applying the conservation of momentum for the system of the 3 balls in the x direction gives:

               v0 = vA prime  +  2 vB (cos 30o)

This is equation 1.

Equation 2 comes from the fact that you are told that the collision is perfectly elastic.  Equating the Kinetic Energy before and after the collision gives:

               ½ m v0^2 =  ½ m (vA prime)^2 +  2( ½ m vB^2 )

This is equation 2.

Solving equation 1 and 2 simultaneously will give vA prime and vB (=vC)

Part b)

If A hits B first, you need to divide the problem into two parts.

First examine the collision of A hitting B. 

Then examine the collision of A hitting C.

For the collision of A hitting B, use the conservation of momentum for the system of A and B. 

vB will still be in the same direction as before (along the line of impact), however, A will have a component in the y direction after the impact in addition to the component in the x direction. So you will get two scalar equations, one in the x direction and the other in the y direction.

Next, obtain an equation using the conservation of KE.

Between these three equations, you can solve for the velocity of B and A after the impact.

Now repeat the exact same procedure for the collision between A and C. 

Apply the conservation of momentum for the system of A and C. 

Then apply the conservation of KE. 

Solve for the final velocities of A and C.

Problem Set #13 – Problem 2

This is another simple, yet tedious problem. 

You are told the initial mass and velocity of a space vehicle.  It explodes. 

You need to realize that during the explosion there are no external forces acting. 

Hence both linear and angular momentum are conserved.

The conservation of linear momentum will give you three scalar equations.

The conservation of angular momentum will give you three scalar equations.

So you have  6 equations and 5 unknowns:  vAx, vAy, vAz, vCy, vC

You can solve this.

Problem Set #13 – Problem 3

Part a)

It will be helpful to reorient the picture and choose an x axis that is in the direction of CD when CD breaks. The y axis is perpendicular to it.

Now consider the system of only A and B and the ring D.

Denote the time just after the cord CD breaks to be t1.

Denote the time when the cord between A and B becomes taut as t2.

At time t1:

A is traveling with a velocity v0 in a negative y direction and theta degrees from the y axis.

B is traveling with a velocity v0 in a negative y direction and theta degrees from the y axis but on the other side of the y axis.

A and B will continue to travel with these velocities until the cord between A and B becomes taut. 

At this point, the ring D will correspond to the center of gravity of the system of A and B. 

(This will be significant when you calculate the linear momentum of the system at t2 in just a moment.)

Between t1 and t2, there are no external forces acting in the plane of motion. 

Hence both linear and angular momentum are conserved.

Calculate the linear momentum of the system at t1.

You should get L = 2m v0 cos(theta) in the negative y direction. 

(The x components cancelled out.)

Now calculate the linear momentum of the system at t2. 

Calculate the linear momentum of the system at t2 by taking the total mass of the system (2m) and multiplying it by the velocity of the mass center, vG. 

At t2, G is the same as D, so vG = vD!

Equating these linear momentums for t1 and t2 will give:

              2m vD = 2m v0 cos(theta)

              vD    = 0.866 v0 in the negative y direction

Part b)

To get the relative speeds at which sphere A and B rotate about D (i.e. vA/D and vB/D) we will use the conservation of angular momentum about the mass center of the system. 

Calculate the angular momentum about point G at time t1. 

(Note that at t1 G is not the same point as D.)

This calculation is relatively straightforward and you should get:

              HG  =            [ rA/G  x  m vA]         +          [ rB/G   x   m vB ]

                      =   [l sin(theta) m v0 sin(theta)]  +  [l sin(theta) m v0 sin(theta) ]

                      =    2 l m v0 sin(theta)^2

Now calculate the angular momentum about point G at time t2:

Note that at t2 G is the same point as D!

In this calculation of angular momentum use the velocities of A and B relative to the frame attached to the mass center, D. 

In other words use vA/D and vB/D instead of the absolute velocities that you used for t1.  (Why can you do this?)

Also noting that vA/D = vB/D you should get:

              HG  =   [ rA  x  m vA/D]   +  [ rB   x   m vB/D ]

                      =        [l m vA/D]       +        [l m vB/D]

                     =   2 l m vA/D

Equating these angular momentums for t1 and t2 will give:

              vA/D = vB/D = 0.25 v0