Final answer:
The velocity recoil of the gun is -72 m/s, and the total energy of the gun and the bullet is 259200 J.
Step-by-step explanation:
The velocity recoil of the gun can be calculated using the law of conservation of momentum. According to this law, the momentum of the bullet and the gun before the firing should be equal to the momentum of the bullet and the gun after the firing.
Given that the mass of the bullet is 10g and the mass of the gun is 10kg, the momentum before the firing is (10g)(0 m/s) + (10kg)(0 m/s) = 0 kg·m/s. Since the bullet embeds in the target after the firing, the momentum after the firing is (10g)(720 m/s) + (10kg)(v), where v is the recoil velocity of the gun.
Equating the two momenta, we have 0 kg·m/s = (10g)(720 m/s) + (10kg)(v), which gives us v = -72 m/s. Therefore, the recoil velocity of the gun is -72 m/s (in the opposite direction of the bullet's velocity).
The total energy of the gun and the bullet can be calculated using the kinetic energy formula. The kinetic energy of the bullet is given by (1/2)(mass of the bullet)(velocity of the bullet)^2. Substituting the values, we get (1/2)(10g)(720 m/s)^2 = 259200 J.
The kinetic energy of the gun is given by (1/2)(mass of the gun)(velocity recoil of the gun)^2. Substituting the values, we get (1/2)(10kg)(-72 m/s)^2 = 259200 J.
Therefore, the total energy of the gun and the bullet is 259200 J.