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Magazine Chapter 8: Problem 25
A hunter on a frozen, essentially frictionless pond uses a rifle that shoots \(4.20 \mathrm{~g}\) bullets at \(965 \mathrm{~m} / \mathrm{s}\). The mass of the hunter (including his gun) is \(72.5 \mathrm{~kg}\). and the hunter holds tight to the gun after firing it. Find the recoil velocity of the hunter if he fires the rifle (a) horizontally and (b) at 56.0 " above the horizontal.
Short Answer
Expert verified
The recoil velocity of the hunter if he fires the rifle (a) horizontally and (b) at 56.0 degrees above the horizontal is approximately \(0.055 \mathrm{~m} / \mathrm{s}\).
Step by step solution
01
Understand the problem and gather data
From the exercise, we have the following known variables: The mass of the bullet (\(m_b)\) is \(4.20 \mathrm{~g}\) or \(0.0042 \mathrm{~kg}\) when converted from grams to kilograms. The velocity of the bullet (\(v_b)\) is \(965 \mathrm{~m} / \mathrm{s}\). The mass of the hunter (\(m_h)\) is \(72.5 \mathrm{~kg}\). The hunter's initial velocity (\(v_h_i)\) is \(0 \mathrm{~m} / \mathrm{s}\) (as he is stationary before the gunshot). The velocity of the hunter after the gunshot (\(v_h_f)\) is what we want to find. The angle 胃 at which the gun is fired in the second scenario is \(56.0^{\circ}\).
02
Apply the conservation of momentum principle in the horizontal direction
According to the principle of conservation of momentum, the total momentum before the gunshot is equal to the total momentum after the gunshot. In the horizontal direction, the total initial momentum (\(m_h \cdot v_{h_i}\) + \(m_b \cdot v_{b_i}\)) is \(72.5 \mathrm{~kg} \times 0 \mathrm{~m} / \mathrm{s}\) + \(0.0042 \mathrm{~kg} \times 0 \mathrm{~m} / \mathrm{s}\) = \(0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). The total final momentum (\(m_h \cdot v_{h_f}\) + \(m_b \cdot v_{b_f}\)) is \(72.5 \mathrm{~kg} \times v_{h_f}\) + \(0.0042 \mathrm{~kg} \times 965 \mathrm{~m} / \mathrm{s}\). Setting these equal gives \(0 = 72.5 \mathrm{~kg} \times v_{h_f} + 0.0042 \mathrm{~kg} \times 965 \mathrm{~m} / \mathrm{s}\). Solving for \(v_{h_f}\), we find \(v_{h_f} = - \frac{0.0042 \mathrm{~kg} \times 965 \mathrm{~m} / \mathrm{s}}{72.5 \mathrm{~kg}}\).
03
Solve for the recoil velocity when the rifle is fired horizontally
Plugging the given values into the equation from step 2, we find \(v_{h_f} = - \frac{0.0042 \mathrm{~kg} \times 965 \mathrm{~m} / \mathrm{s}}{72.5 \mathrm{~kg}} = -0.055132 \mathrm{~m} \mathrm{s}^{-1}\). The negative sign indicates that the hunter moves in the opposite direction of the bullet, which is expected. Therefore, the recoil velocity of the hunter when he fires the rifle horizontally is \(0.055 \mathrm{~m} / \mathrm{s}\) (approximately, when rounded to three decimal places).
04
Apply the conservation of momentum principle in the vertical direction
In the vertical direction, assuming that upward is positive, the total initial momentum is \(0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\) because neither the hunter nor the bullet were moving vertically before the gunshot. The final momentum is \(m_h \cdot v_{h_{f_{vert}}} + m_b \cdot v_{b_{f_{vert}}}\), where \(v_{h_{f_{vert}}}\) is the vertical component of the hunter鈥檚 final velocity and \(v_{b_{f_{vert}}}\) is the vertical component of the bullet鈥檚 final velocity. For the bullet, \(v_{b_{f_{vert}}} = 965 \mathrm{~m} / \mathrm{s} \times \sin(56^{\circ})\). Setting the initial and final momentum equal gives \(0 = 72.5 \mathrm{~kg} \times v_{h_{f_{vert}}} + 0.0042 \mathrm{~kg} \times 965 \mathrm{~m} / \mathrm{s} \times \sin(56^{\circ})\). Solving for \(v_{h_{f_{vert}}}\), we find \(v_{h_{f_{vert}}} = - \frac{0.0042 \mathrm{~kg} \times 965 \mathrm{~m} / \mathrm{s} \times \sin(56^{\circ})}{72.5 \mathrm{~kg}}\).
05
Solve for the recoil velocity when the rifle is fired at 56 degrees above horizontal
Plugging the given values into the equation from step 4, we find \(v_{h_{f_{vert}}} = - \frac{0.0042 \mathrm{~kg} \times 965 \mathrm{~m} / \mathrm{s} \times \sin(56^{\circ})}{72.5 \mathrm{~kg}} = -0.052 \mathrm{~m} \mathrm{s}^{-1}\). The net final velocity of the hunter is the vector sum of the horizontal and vertical components. The magnitude of this vector will be the recoil velocity when the rifle is fired at 56 degrees above the horizontal. Given that the horizontal and vertical components are equal in magnitude and opposite in direction, the recoil velocity when the rifle is fired at 56 degrees above the horizontal will remain the same, \(0.055 \mathrm{~m} / \mathrm{s}\) (approximately).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conservation of Momentum
The principle of conservation of momentum is a fundamental concept in physics stating that the total momentum of a closed system remains constant if no external forces are acting upon it. Momentum, a measure of the amount of motion an object has, is calculated as the product of an object's mass and velocity (p = mv).
When a hunter fires a bullet from a rifle, the bullet gains forward momentum while the hunter and rifle gain an equal amount of momentum in the opposite direction. This interaction illustrates conservation of momentum鈥攖he momentum before firing (which is zero, as both hunter and bullet are stationary) is equal to the total momentum after the bullet is fired.
To calculate the recoil velocity of the hunter, we set the total initial momentum equal to the total final momentum. Since there are no external horizontal forces, the horizontal momentum before the shot must be equal to the horizontal momentum after. This relationship allows us to solve for the unknown velocity of the hunter after firing the rifle. It's important to note that since momentum has a direction, recoil velocity is negative, indicating the direction is opposite to that of the bullet.
When a hunter fires a bullet from a rifle, the bullet gains forward momentum while the hunter and rifle gain an equal amount of momentum in the opposite direction. This interaction illustrates conservation of momentum鈥攖he momentum before firing (which is zero, as both hunter and bullet are stationary) is equal to the total momentum after the bullet is fired.
To calculate the recoil velocity of the hunter, we set the total initial momentum equal to the total final momentum. Since there are no external horizontal forces, the horizontal momentum before the shot must be equal to the horizontal momentum after. This relationship allows us to solve for the unknown velocity of the hunter after firing the rifle. It's important to note that since momentum has a direction, recoil velocity is negative, indicating the direction is opposite to that of the bullet.
Projectile Motion
Projectile motion describes the motion of an object that is launched into the air and is subject to gravitational acceleration but no other forces. The motion of a projectile is two-dimensional and can be broken down into horizontal and vertical components that are independent of each other.
The horizontal component (vx) involves constant velocity motion, as there are no horizontal forces once the projectile is in the air (ignoring air resistance). The vertical component (vy) involves acceleration due to gravity, which is downward. For the firing of the rifle, the projectile (the bullet) has both a horizontal and a vertical component when shot at an angle.
The vertical motion affects the vertical recoil velocity of the hunter. By solving for the vertical component of the momentum post-shot, as done in the step-by-step solution, one can determine how the vertical forces play a role in the overall motion of the hunter. Understanding projectile motion is essential for predicting the trajectory of the bullet and the recoil path of the shooter.
The horizontal component (vx) involves constant velocity motion, as there are no horizontal forces once the projectile is in the air (ignoring air resistance). The vertical component (vy) involves acceleration due to gravity, which is downward. For the firing of the rifle, the projectile (the bullet) has both a horizontal and a vertical component when shot at an angle.
The vertical motion affects the vertical recoil velocity of the hunter. By solving for the vertical component of the momentum post-shot, as done in the step-by-step solution, one can determine how the vertical forces play a role in the overall motion of the hunter. Understanding projectile motion is essential for predicting the trajectory of the bullet and the recoil path of the shooter.
Momentum in Two Dimensions
When dealing with scenarios such as a bullet being fired from a rifle, it's essential to consider momentum in two dimensions. The motion of objects can be analyzed separately along the horizontal and vertical axes. This concept, known as the independence of motion, is crucial as it simplifies calculations and understanding of physical phenomena.
In our hunter's scenario, when the rifle is fired at an angle above the horizontal, the bullet's final velocity has a horizontal and a vertical component. Similarly, the hunter's recoil velocity also has two components. Since the bullet is fired at an angle, we use trigonometry to calculate the vertical component of the bullet's velocity, which in turn allows us to find the vertical component of the hunter's recoil velocity using conservation of momentum.
The conservation of momentum in two dimensions ensures that momentum is conserved separately in each direction. Therefore, to find the total recoil velocity when the rifle is fired at an angle, we must account for both the horizontal and vertical components. The hunter's final recoil velocity is the vector sum of the two components, illustrating how two-dimensional momentum analysis is applied to real-world problems.
In our hunter's scenario, when the rifle is fired at an angle above the horizontal, the bullet's final velocity has a horizontal and a vertical component. Similarly, the hunter's recoil velocity also has two components. Since the bullet is fired at an angle, we use trigonometry to calculate the vertical component of the bullet's velocity, which in turn allows us to find the vertical component of the hunter's recoil velocity using conservation of momentum.
The conservation of momentum in two dimensions ensures that momentum is conserved separately in each direction. Therefore, to find the total recoil velocity when the rifle is fired at an angle, we must account for both the horizontal and vertical components. The hunter's final recoil velocity is the vector sum of the two components, illustrating how two-dimensional momentum analysis is applied to real-world problems.
