We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
Let's start with energy conservation law:
![\frac{mv^2}{2}+\frac{2mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+\frac{2mv_{B}^2}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E2%7D%7B2%7D%2B%5Cfrac%7B2mv%5E2%7D%7B2%7D%2B0.75E%3D%5Cfrac%7Bmv_%7BA%7D%5E2%7D%7B2%7D%2B%5Cfrac%7B2mv_%7BB%7D%5E2%7D%7B2%7D)
![\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2](https://tex.z-dn.net/?f=%5Cfrac%7B3mv%5E2%7D%7B2%7D%2B0.75E%3D%5Cfrac%7Bmv_%7BA%7D%5E2%7D%7B2%7D%2Bmv_%7BB%7D%5E2)
This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
![mv+2mv=mv_A+2mv_B\\ 3mv=mv_A+2mv_B\\](https://tex.z-dn.net/?f=mv%2B2mv%3Dmv_A%2B2mv_B%5C%5C%203mv%3Dmv_A%2B2mv_B%5C%5C%20)
Because both carts have the same mass we can cancel those out:
![3v=v_A+2v_B](https://tex.z-dn.net/?f=3v%3Dv_A%2B2v_B)
Now we have our system of equation that we have to solve:
Part A
We need to solve our system for
![v_a](https://tex.z-dn.net/?f=%20v_a)
. We will solve second equation for
![v_b](https://tex.z-dn.net/?f=v_b)
and then plug that in the first equation.
![3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=\frac{3v-v_A}{2}](https://tex.z-dn.net/?f=3v%3Dv_A%2B2v_B%5C%5C%203v-v_A%3D2v_B%5C%5C%20v_B%3D%5Cfrac%7B3v-v_A%7D%7B2%7D)
Now we have to plug this in the first equation:
![\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\](https://tex.z-dn.net/?f=%5Cfrac%7B3mv%5E2%7D%7B2%7D%2B0.75E%3D%5Cfrac%7Bmv_%7BA%7D%5E2%7D%7B2%7D%2Bmv_%7BB%7D%5E2%5C%5Cv_B%3D%5Cfrac%7B3v-v_A%7D%7B2%7D%5C%5C%20)
We will multiply the first equation with 2 and divide by m:
![3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\](https://tex.z-dn.net/?f=3v%5E2%2B%5Cfrac%7B3E%7D%7B2m%7D%3Dv_%7BA%7D%5E2%7D%2B2v_%7BB%7D%5E2%5C%5Cv_B%3D%5Cfrac%7B3v-v_A%7D%7B2%7D%5C%5C%20)
Now we plug in the second equation into first one:
![3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+2\frac{(3v-v_A)^2}{4}\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+\frac{9v^2-6v\cdot v_A+v_{A}^2}{2} /\cdot 2\\ 6v^2+\frac{3E}{m}=2v_{A}^2+9v^2-6v\cdot v_A+v_{A}^2}\\ 3v_A^2-6v\cdot v_a+3(v^2-\frac{E}{m})=0/\cdot\frac{1}{3}\\ v_A^2-3v\cdot v_A+ (v^2-\frac{E}{m})=0](https://tex.z-dn.net/?f=3v%5E2%2B%5Cfrac%7B3E%7D%7B2m%7D%3Dv_%7BA%7D%5E2%7D%2B2v_%7BB%7D%5E2%5C%5C%203v%5E2%2B%5Cfrac%7B3E%7D%7B2m%7D%3Dv_%7BA%7D%5E2%7D%2B2%5Cfrac%7B%283v-v_A%29%5E2%7D%7B4%7D%5C%5C%203v%5E2%2B%5Cfrac%7B3E%7D%7B2m%7D%3Dv_%7BA%7D%5E2%7D%2B%5Cfrac%7B9v%5E2-6v%5Ccdot%20v_A%2Bv_%7BA%7D%5E2%7D%7B2%7D%20%2F%5Ccdot%202%5C%5C%206v%5E2%2B%5Cfrac%7B3E%7D%7Bm%7D%3D2v_%7BA%7D%5E2%2B9v%5E2-6v%5Ccdot%20v_A%2Bv_%7BA%7D%5E2%7D%5C%5C%203v_A%5E2-6v%5Ccdot%20v_a%2B3%28v%5E2-%5Cfrac%7BE%7D%7Bm%7D%29%3D0%2F%5Ccdot%5Cfrac%7B1%7D%7B3%7D%5C%5C%20v_A%5E2-3v%5Ccdot%20v_A%2B%20%28v%5E2-%5Cfrac%7BE%7D%7Bm%7D%29%3D0%20)
We end up with quadratic equation that we have to solve, I won't solve it by hand.
Coefficients are:
![a=1\\ b=-6v\\ c=v^2-\frac{E}{m}](https://tex.z-dn.net/?f=a%3D1%5C%5C%0Ab%3D-6v%5C%5C%0Ac%3Dv%5E2-%5Cfrac%7BE%7D%7Bm%7D)
Solutions are:
Part B
We do the same thing here, but we must express
![v_a](https://tex.z-dn.net/?f=%20v_a)
from momentum equation:
![3v=v_A+2v_B\\ v_A=3v-2v_B](https://tex.z-dn.net/?f=3v%3Dv_A%2B2v_B%5C%5C%0Av_A%3D3v-2v_B)
Now we plug this into our energy conservation equation:
![3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_A={3v-v_B}\\ 3v^2+\frac{3E}{2m}=(3v-v_B)^2+2v_B^2\\ 3v^2+\frac{3E}{2m}=9v^2-6v\cdot v_B+v_B^2+2v_B^2\\ 3v^2+\frac{3E}{2m}=3v_B^2-6v\cdot v_B+9v^2\\ 3v_B^2-6v\cdot v_B+9v^2-3v^2-\frac{3E}{2m}=0\\ 3v_B^2-6v\cdot v_B+(6v^2-\frac{3E}{2m})=0 ](https://tex.z-dn.net/?f=3v%5E2%2B%5Cfrac%7B3E%7D%7B2m%7D%3Dv_%7BA%7D%5E2%7D%2B2v_%7BB%7D%5E2%5C%5Cv_A%3D%7B3v-v_B%7D%5C%5C%0A3v%5E2%2B%5Cfrac%7B3E%7D%7B2m%7D%3D%283v-v_B%29%5E2%2B2v_B%5E2%5C%5C%0A3v%5E2%2B%5Cfrac%7B3E%7D%7B2m%7D%3D9v%5E2-6v%5Ccdot%20v_B%2Bv_B%5E2%2B2v_B%5E2%5C%5C%0A3v%5E2%2B%5Cfrac%7B3E%7D%7B2m%7D%3D3v_B%5E2-6v%5Ccdot%20v_B%2B9v%5E2%5C%5C%0A3v_B%5E2-6v%5Ccdot%20v_B%2B9v%5E2-3v%5E2-%5Cfrac%7B3E%7D%7B2m%7D%3D0%5C%5C%0A3v_B%5E2-6v%5Ccdot%20v_B%2B%286v%5E2-%5Cfrac%7B3E%7D%7B2m%7D%29%3D0%0A)
Again we end up with quadratic equation. Coefficients are:
![a=3\\ b=-6v\\ c=6v^2-\frac{3E}{2m}](https://tex.z-dn.net/?f=a%3D3%5C%5C%0Ab%3D-6v%5C%5C%0Ac%3D6v%5E2-%5Cfrac%7B3E%7D%7B2m%7D)
Solutions are: