Final answer:
The magnitude of the recoil velocity of the hunter is 2.18 m/s.
Step-by-step explanation:
When a hunter fires a rifle, the bullet and rifle experience a recoil effect due to the conservation of momentum. According to the law of conservation of momentum, the vector sum of the momentum of the bullet and rifle before firing should be equal to the vector sum of the momentum of the bullet and rifle after firing.
To find the magnitude of the recoil velocity of the hunter, we need to consider the momentum of the bullet and the rifle separately. The magnitude of the initial momentum of the bullet is given by:
Initial momentum of bullet = mass of bullet * initial velocity of bullet
Substituting the given values, we find:
Initial momentum of bullet = (0.005 kg) * (1075 m/s) = 5.375 kg·m/s
The momentum of the rifle and hunter system before firing is zero since they were at rest. Therefore, according to the conservation of momentum, the magnitude of the momentum of the bullet and rifle system after firing should also be 5.375 kg·m/s.
Now, let's consider the momentum of the bullet and rifle system after firing at an angle of 65.0° above the horizontal. We can find the horizontal and vertical components of the momentum using trigonometry. The horizontal component of the momentum is given by:
Horizontal component of momentum = magnitude of momentum * cos(angle)
Substituting the values, we find:
Horizontal component of momentum = (5.375 kg·m/s) * cos(65.0°) ≈ 2.18 kg·m/s
Similarly, the vertical component of the momentum is given by:
Vertical component of momentum = magnitude of momentum * sin(angle)
Substituting the values, we find:
Vertical component of momentum = (5.375 kg·m/s) * sin(65.0°) ≈ 4.90 kg·m/s
Since the hunter holds tightly to the gun after firing it, the magnitude of the recoil velocity of the hunter will be equal to the magnitude of the horizontal component of the momentum:
Magnitude of recoil velocity = horizontal component of momentum = 2.18 m/s