Question

Cart A of inertia m has attached to its front end a device that explodes when it hits anything, releasing a quantity of energy E. This cart is moving to the right with speed v when it collides head-on with cart B of inertia 2m traveling to the left at the same speed v. The explosive goes off when the carts hit, causing them to rebound from each other. The initial direction of motion of cart A (to the right) is the +x direction.

Part A
If one-quarter of the explosive energy is dissipated into the incoherent energy of noise and deformation of the carts, what is the final velocity of cart A?
Express your answer in terms of the variables E, v, and m.

Part B
If one-quarter of the explosive energy is dissipated into the incoherent energy of noise and deformation of the carts, what is the final velocity of cart B?
Express your answer in terms of the variables E, v, and m.

Answer 1

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:

`(mv^2)/(2)+(2mv^2)/(2)+0.75E=(mv_(A)^2)/(2)+(2mv_(B)^2)/(2)`

`(3mv^2)/(2)+0.75E=(mv_(A)^2)/(2)+mv_(B)^2`
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\\`
Because both carts have the same mass we can cancel those out:

`3v=v_A+2v_B`
Now we have our system of equation that we have to solve:

`(3mv^2)/(2)+0.75E=(mv_(A)^2)/(2)+mv_(B)^2\\ 3v=v_A+2v_B`
Part A
We need to solve our system for
`v_a`. We will solve second equation for
`v_b` and then plug that in the first equation.

`3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=(3v-v_A)/(2)`
Now we have to plug this in the first equation:

`(3mv^2)/(2)+0.75E=(mv_(A)^2)/(2)+mv_(B)^2\\v_B=(3v-v_A)/(2)\\`
We will multiply the first equation with 2 and divide by m:

`3v^2+(3E)/(2m)=v_(A)^2}+2v_(B)^2\\v_B=(3v-v_A)/(2)\\`
Now we plug in the second equation into first one:

`3v^2+(3E)/(2m)=v_(A)^2}+2v_(B)^2\\ 3v^2+(3E)/(2m)=v_(A)^2}+2((3v-v_A)^2)/(4)\\ 3v^2+(3E)/(2m)=v_(A)^2}+(9v^2-6v\cdot v_A+v_(A)^2)/(2) /\cdot 2\\ 6v^2+(3E)/(m)=2v_(A)^2+9v^2-6v\cdot v_A+v_(A)^2}\\ 3v_A^2-6v\cdot v_a+3(v^2-(E)/(m))=0/\cdot(1)/(3)\\ v_A^2-3v\cdot v_A+ (v^2-(E)/(m))=0`
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-(E)/(m)`
Solutions are:

`v_A=\frac{3v+\sqrt{5v^2+(4E)/(m)}}{2},\:v_A=\frac{3v-\sqrt{5v^2+(4E)/(m)}}{2}`
Part B
We do the same thing here, but we must express
`v_a` from momentum equation:

`3v=v_A+2v_B\\ v_A=3v-2v_B`
Now we plug this into our energy conservation equation:

`3v^2+(3E)/(2m)=v_(A)^2}+2v_(B)^2\\v_A={3v-v_B}\\ 3v^2+(3E)/(2m)=(3v-v_B)^2+2v_B^2\\ 3v^2+(3E)/(2m)=9v^2-6v\cdot v_B+v_B^2+2v_B^2\\ 3v^2+(3E)/(2m)=3v_B^2-6v\cdot v_B+9v^2\\ 3v_B^2-6v\cdot v_B+9v^2-3v^2-(3E)/(2m)=0\\ 3v_B^2-6v\cdot v_B+(6v^2-(3E)/(2m))=0`
Again we end up with quadratic equation. Coefficients are:

`a=3\\ b=-6v\\ c=6v^2-(3E)/(2m)`
Solutions are:

`v_B=\frac{6v+\sqrt{-36v^2+(18E)/(m)}}{6},\:v_B=\frac{6v-\sqrt{-36v^2+(18E)/(m)}}{6}`