Finding velocities in super elastic collision

[jcruise322]

Homework Statement

Two 1.0-kg carts are coupled together and placed on a very long
horizontal track that is at rest in the Earth frame of reference. The combination is launched so that at t
= 0 it is moving to the right at speed 2.0 m/s in the Earth frame, with cart #1 in front and cart #2 in
back. The coupling contains an explosive charge that is remotely ignited at time t1 > 0, releasing 18 J
of energy. Half of this released energy is dissipated into noise, thermal energy, and minor damage to
each cart. Assume zero friction between the carts and the track.

What is the final velocity (after the explosion, time t > t1) of
cart #2 in the center-of-mass (CM of the 2 cart system) reference frame?

The answer is 3m/s to the right

Homework Equations

KE=.5*m*v^2
m1*v1+m2*v2=m1*v3+m1*v4
COM velocity: V1*m1+V2+m2/(m2+m1)

The Attempt at a Solution

[/B]
I already found the center of mass velocity, 2m/s.
So...we assume 9 joules of energy was transformed into kinetic energy.
So, energy of system is given by: PE+KEinitial=KE final
so: 9 joules+4 joules from both cartes=13 joules of kinetic energy final: 13=.5mv3^2+.5*mv4^2
MV is conserved: m1v1+m2v2=m1v3+m1v4
Since all the M's are the same and the velocity is 2m/s for both:
4=v3+v4

v2=(4-v1). Plug into kinetic energy equation.

26=2v^2-8v=16

v=-1, 5. Since both masses are the same, I assume 5m/s is the velocity of both carts after the collision.
Now comes the tricky part...do I assume the center of mass velocity stays the same? If so, I would subtract 2 and get the book's answer 3. How come the center of mass velocity stays the same after this collision? I thought this was only true for elastic collisions...do we treat this bomb like an elastic collision in this case, and if so, why?

Regards,

JT

Any help would be appreciated :)

 

[haruspex]

v=-1, 5.

In the ground reference frame, yes.

Since both masses are the same, I assume 5m/s is the velocity of both carts after the collision.

Why would you assume that? They are not moving in the same direction at the same speed, surely.

Now comes the tricky part...do I assume the center of mass velocity stays the same? If so, I would subtract 2 and get the book's answer 3. How come the center of mass velocity stays the same after this collision? I thought this was only true for elastic collisions...do we treat this bomb like an elastic collision in this case, and if so, why?

What external forces act in the horizontal direction on the two-cart-plus-charge system?
By the way, the answer you quote from the book is wrong. Cart 2, the left-hand cart, will be moving to the left relative to the common mass centre.

 

[jcruise322]

I just checked, yes it is to the left. I wrote it down correctly. And there are no external forces acting on the cart1/cart2 system..
Oh wow, yeah I get it.
5 for V1, -1 for v2. The center of mass reference frame is 2m/s, so the center of mass is moving at 2m/s to the right, and m2 is 1m/s to the left, so, 3m/s to left in the center of mass reference frame.
Thanks Haruspex! I will probably be posting a few more questions later

 

[ehild]

The KE of the system is the sum of the kinetic energies relative to the center of mass and the KE of the CM: If the velocities with respect to the CM are u1 and u2, and the velocity of the CM is V
$KE=0.5 (m_1 u_1^2+m_2 u_2^2)+0.5(m_1+m_2)V^2$
Without external force, the velocity of the center of mass does not change. The available energy goes to the relative kinetic energy of the masses in the center of mass frame.
$0.5 (m_1 u_1^2+m_2 u_2^2)=9$
$m_1u_1+m_2u_2=0$
$m_1=m_2=1$, so $u_2= -u_1$ and $u_1^2=9$.