Recoil kinetic energy in gun and bullet system

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cheez
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You forget to hold your gun against your shoulder and it recoils into you. How much Joules of kinetic energy end up in your shoulder? Assume you and the gun move together in the end.
masses: You=90 kg, gun=10kg, bullet= 0.0085kg
velocity of the bullet = 87m/s

I read the answer of it, but I don't get why the answer is that.
And the answer is

1/2 M gun x (M bullet * V bullet / M gun)^2
- 1/2 M gun+you x (M bullet * V bullet/ M gun+you) ^2thx!
 
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What have you tried?

After the impact, energy is conserved (you can't assume conservation during the impact though).
 
Päällikkö said:
What have you tried?

After the impact, energy is conserved (you can't assume conservation during the impact though).

Before I checked the answer, I calculated the recoil speed of the gun. And I set the equation using conservation of momentum to find the velocity of "you+gun".
i.e (M bullet * V bullet / M gun) * M gun = M you+gun V you+gun

But I don't know why the final answer is the Kinetic energy from the gun after the bullet is fired minus the Kinetic energy of "you+gun" after the impact.
 
By conservation of momentum, you have calculated the the momentum of the system directly after the after this inelastic impact. By looking at the kinetic energy before & after the collision you will notice that energy is lost somewhere; ask yourself where this energy has gone.

You also have to assume that no energy is lost during the time it take the rifle to you.

Regards,
Sam