To solve this problem, we can apply the law of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the initial velocity of the soldier as V1 (which we need to find), the final velocity of the soldier as V2, the mass of the bullet as m1 (5 grams or 0.005 kg), and the mass of the soldier as m2 (70 kg). The initial velocity of the bullet is given as 680 m/s.
Before the collision: Total momentum = m1 * V1 + m2 * 0 (since the soldier is initially at rest)
After the collision: Total momentum = (m1 + m2) * V2
According to the law of conservation of momentum: m1 * V1 + m2 * 0 = (m1 + m2) * V2
Now we can solve for V1 to find the velocity with which the soldier is thrown backwards.
m1 * V1 = (m1 + m2) * V2 0.005 kg * V1 = (0.005 kg + 70 kg) * V2 0.005 kg * V1 = 70.005 kg * V2
Dividing both sides by 0.005 kg: V1 = 70.005 kg * V2 / 0.005 kg V1 = 14,001 * V2
Now we need to find V2, the final velocity of the soldier. Since the bullet remains within the victim, we can assume an inelastic collision. In this type of collision, the two objects stick together after the collision, and their final velocity can be calculated using the conservation of momentum and conservation of energy.
Conservation of momentum: m1 * V1 + m2 * 0 = (m1 + m2) * V2
Conservation of energy: (1/2) * m1 * V1^2 + (1/2) * m2 * 0^2 = (1/2) * (m1 + m2) * V2^2
Since the bullet is small compared to the soldier, we can neglect its kinetic energy in the initial state (0.5 * m1 * V1^2). We are left with:
0 = (1/2) * (m1 + m2) * V2^2
Simplifying further: 0 = (0.5 kg) * (0.005 kg + 70 kg) * V2^2 0 = 35.0025 * V2^2
Since V2 cannot be zero (as the soldier moves), we can solve for V2:
V2^2 = 0 / 35.0025 V2 = 0 m/s
Therefore, the final velocity of the soldier (V2) is 0 m/s.
Substituting this result back into the equation for V1: V1 = 14,001 * V2 V1 = 14,001 * 0 V1 = 0 m/s
Hence, the soldier is not thrown backwards and remains at rest after being hit by the bullet.