Two carts are held together, and when released they are pushed apart by a spring

[TjGrinnell]

Homework Statement

Two carts move together at 1.3 m/s north (in the y direction) along a frictionless track. The front cart(Cart A) has a mass of 750 g while the back cart(Cart B) has a mass of 300 g. At t = 0 the carts are at y = 0. At t = 1.00s a spring pushes the two carts apart, releasing 0.25 J of spring energy.
What is the total kinetic energy of the system after the spring releases?
What are the velocities of each cart after the spring releases?
Where is the back cart at t = 4.00 s?

Relevant Equations

KE=1/2mv^2
v=sqrt(K/(1/2m))

Ever since the switch to online learning, I have been having trouble with understanding the topics right out of the book. And so I am just not sure if I am ever doing anything right in physics. So far I have calculated the KE of both carts KE(A)=0.634 J and KE(B)=0.254 J. I am unsure how exactly the push at t= 1.00s will affect the carts. Cart A go forward and Cart B will go backwards obviously, but would I simply add the work from the spring to the carts? As for calculating the velocities, Cart A would be moving faster north after the interaction and Cart B would begin to move south, so what would I need to do to calculate their velocity?

 

[etotheipi]

Focus your attention to the instances right before and right after the spring releases. You have two conserved quantities: total energy, and momentum.

Let the components of velocity in the $y$ direction after the spring releases be $v_a$ and $v_b$; these might be positive or negative.

 

[TjGrinnell]

Focus your attention to the instances right before and right after the spring releases. You have two conserved quantities: total energy, and momentum.

So would the conservation equation look something like this? KE(A_i)+KE(B_i)+W=KE(A_f)+KE(B_f). I am especially bad at doing these since the book uses different symbols than what we started to use in class, and I have never been able to fully grasp how to set them up.

 

[etotheipi]

So would the conservation equation look something like this? KE(A_i)+KE(B_i)+W=KE(A_f)+KE(B_f). I am especially bad at doing these since the book uses different symbols than what we started to use in class, and I have never been able to fully grasp how to set them up.

That looks fine, I personally would use $\text{U}_{EPE}$ instead of $W$, however the work done by the spring on the two carts equals the initial $U_{EPE}$ here (the final being zero) so it doesn't make any difference.

An easy way to think about it is to define your system (e.g. here, the two carts and the spring), and set the transfers of energy into the system equal to the change in the total energy of the system. The carts-spring system is closed (no external work), so we get

${\text{KE}_A}_1 + {\text{KE}_B}_1 + \text{U}_{EPE} = {\text{KE}_A}_2 + {\text{KE}_B}_2$

 

[TjGrinnell]

That looks fine, I personally would use $\text{U}_{EPE}$ instead of $W$, however the work done by the spring on the two carts equals the initial $U_{EPE}$ here (the final being zero) so it doesn't make any difference.

An easy way to think about it is to define your system (e.g. here, the two carts and the spring), and set the transfers of energy into the system equal to the change in the total energy of the system. The carts-spring system is closed (no external work), so we get

${\text{KE}_A}_1 + {\text{KE}_B}_1 + \text{U}_{EPE} = {\text{KE}_A}_2 + {\text{KE}_B}_2$

Would Total Kinetic Energy be 0.634+0.254+0.25=1.14J? And what about the velocities afterwards, 1.14=1/2(.75)v_Af²+1/2(.3)v_Bf² how do I go about splitting up this equation so solve for each variable?

 

[scottdave]

Do you know how to account for Conservation of Momentum? The spring did not change the momentum. So the cart pair has the same momentum before spring release as after. Momentum is a vector quantity with direction.

You now can use this equation with the energy equation to solve for the 2 unknowns.

 

[TjGrinnell]

Do you know how to account for Conservation of Momentum? The spring did not change the momentum. So the cart pair has the same momentum before spring release as after. Momentum is a vector quantity with direction.

You now can use this equation with the energy equation to solve for the 2 unknowns.

I think I got it after some more pondering, brain is fried after trying to study for all these online finals. Thank you for the reply though.

 

[scottdave]

Good Luck. Hopefully you've learned a few things