Hi! Thank you very much-- I know this is a 2dimensional inelastic collision but i'm not sure how to solve--

"The mass of a blue block is 20% greater than the mass of the green block. Before colliding, the blocks approach each ther with momenta of equal magnitudes and opposite directions, and the green block has an initial speed of 10 m/s. Find the speeds of the blocks after the collision if half the kinetic energy is lost during the collision" There is a picture illustrating the collision---the green block (G) is moving to the right along the x axis, the blue block (B) is moving to the left along the x axis;;after the collision, G is moving 30 degrees north of east, and B is moving 30 degrees south of west.

I know that since initial momenta are equal, (with G = mass of green block)

(G)(10 m/s) = (1.2 G)(V of blue block)
divide both sides by G
so I solved for V of blue block = 8.333 m/s

Since "half the kinetic energy is lost during the collision"
I think
KEi(G) + KE(ib) = 0.5[KEf(G) + KEf(B)]

But I DO NOT know how to do the in between steps---I'm not sure how to work with velocity components in this problem!

Please assist! Thank you very much!!

To solve this 2-dimensional inelastic collision problem, we can break down the velocities of the blocks into their x- and y-components.

Let's start by assigning some variables:
- Let mG be the mass of the green block.
- The mass of the blue block is 20% greater, so its mass is 1.2 times mG (1.2G).
- The initial velocity of the green block (before the collision) is 10 m/s.
- The initial velocity of the blue block (before the collision) is unknown but can be represented as VBx since it is moving along the x-axis.

Given that half the kinetic energy is lost during the collision, we can set up an equation using the conservation of kinetic energy:
KEi(G) + KEi(B) = 0.5[KEf(G) + KEf(B)]

Let's work through the steps to solve the problem:

1. Determine the initial momentum equation:
Before the collision, the blocks approach each other with equal magnitudes of momentum in opposite directions.
Therefore, the initial momentum equation can be written as:
mG * 10 m/s = (1.2G) * VBx

2. Solve for VBx:
Divide both sides of the equation by mG:
10 m/s = 1.2 * VBx

VBx = 10 m/s / 1.2
VBx = 8.333 m/s

3. Express the velocities in terms of their components:
After the collision, both blocks have velocities with x- and y-components.

Let VGf and VBf represent the final velocities of the green and blue blocks, respectively.
We are given some information about their angles:
- The green block moves 30 degrees north of east (above the x-axis).
- The blue block moves 30 degrees south of west (below the x-axis).

4. Break down the final velocities into their x- and y-components:
To find the components, we can use trigonometry.

VGf = VGfx + VGfy:
- The x-component of the green block's velocity is VGfx = VGf * cos(30°).
- The y-component of the green block's velocity is VGfy = VGf * sin(30°).

VBf = VBfx + VBfy:
- The x-component of the blue block's velocity is VBfx = VBf * cos(30°).
- The y-component of the blue block's velocity is VBfy = VBf * sin(30°).

5. Solve for the final velocities using the fact that momentum is conserved:
Since momentum is conserved, the equation can be written as:
mG * 10 m/s = mG * VGfx + (1.2G) * VBfx

We can substitute the previously found value for VBx and rearrange the equation to solve for VGfx:
10 m/s = VGfx * mG + 8.333 m/s * 1.2 * mG

Solve for VGfx:
VGfx = (10 m/s - 8.333 m/s * 1.2 * mG) / mG

Similarly, substitute VBx into the equation and solve for VBfx.

Once you have the x-components of the final velocities, you can find the y-components using the trigonometric equations mentioned in step 4.