A girl holds a 1.7 kg air rifle loosely and fires a bullet of mass 2 g. The muzzle velocity of the bullet is 160 m/s. What is the magnitude of the recoil velocity of the gun?
Since total momentum is conserved (and remains zero) the forward mometum of the bullet equals the the momentum of the gun in the opposite direction.
Holding the rifle tightly will diminish the recoil velocity, by applying an outside force. That is why the word "loosely" was added to the question.
To find the magnitude of the recoil velocity of the gun, we can use the principle of conservation of momentum. According to this principle, the total momentum before the firing of the bullet is equal to the total momentum after the firing.
The momentum of an object is calculated by multiplying its mass by its velocity.
Let's denote the recoil velocity of the gun as V_gun, the velocity of the bullet as V_bullet, and the mass of the gun as M_gun.
Before the firing:
M_gun * 0 (since the gun is at rest) = 1.7 kg * V_gun
After the firing:
2 g * 160 m/s = (1.7 kg + 0.002 kg) * V_gun
First, we need to convert the mass of the bullet from grams to kilograms:
2 g = 0.002 kg
Now, we can simplify the equation:
0 = (1.702 kg) * V_gun - (0.002 kg) * (160 m/s)
The equation further simplifies to:
0 = (1.702 kg) * V_gun - 0.32 kg m/s
We can isolate V_gun by moving the term with V_gun to one side of the equation:
(1.702 kg) * V_gun = 0.32 kg m/s
Finally, divide both sides of the equation by (1.702 kg) to solve for V_gun:
V_gun = 0.32 kg m/s / 1.702 kg
Simplifying this expression gives us the magnitude of the recoil velocity of the gun.