Problem 25 - University Physics with Modern Physics

Question

A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20 -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 \(56.0^{\circ}\) above the horizontal.

Step-by-Step Solution

Answer

(a) Recoil velocity is -0.056 m/s for horizontal firing. (b) Recoil velocity is -0.031 m/s for 56° angle firing.

Step 1: Analyze the Problem

We need to find the recoil velocity of the hunter when he fires a gun, once horizontally and once at an angle. Using the principle of conservation of momentum, we can solve this by setting the momentum before and after firing as equal since there is no external force.

Step 2: Define the Variables

Let the mass of the bullet be \(m_b = 4.20\, \mathrm{g} = 0.0042\, \mathrm{kg}\) and its speed \(v_b = 965\, \mathrm{m/s}\). The mass of the hunter including the gun is \(m_h = 72.5\, \mathrm{kg}\). The unknowns are the recoil velocities \(v_{hx}\) for horizontal firing and \(v_{hr}\) for firing at an angle.

Step 3: Apply Conservation of Momentum for Horizontal Firing

Before the gun is fired, the total momentum is zero. When the bullet is fired horizontally, \(m_b \times v_b + m_h \times v_{hx} = 0\). So, \(v_{hx} = - \frac{m_b \times v_b}{m_h}\). Substitute values to find \(v_{hx}\).

Step 4: Calculate Horizontal Recoil Velocity

Substitute the values into the equation:\[v_{hx} = - \frac{0.0042\, \mathrm{kg} \times 965\, \mathrm{m/s}}{72.5\, \mathrm{kg}}\]\[v_{hx} = -0.056\, \mathrm{m/s}\]The negative sign indicates the direction opposite to the bullet.

Step 5: Apply Conservation of Momentum for Firing at an Angle

For firing at an angle, only the horizontal component of the bullet's velocity contributes to the recoil. The horizontal component is \(v_{b} \times \cos(56.0^{\circ})\). Momentum conservation equation: \(m_b \times v_b \times \cos(56.0^{\circ}) + m_h \times v_{hr} = 0\). Solve for \(v_{hr}\).

Step 6: Calculate Recoil Velocity for Firing at an Angle

Substitute the values:\[v_{hr} = - \frac{0.0042\, \mathrm{kg} \times 965\, \mathrm{m/s} \times \cos(56.0^{\circ})}{72.5\, \mathrm{kg}}\]\[v_{hr} = -0.031\, \mathrm{m/s}\]Again, the negative indicates recoil opposite to bullet direction.

Key Concepts

Recoil Velocity

Recoil velocity is the speed at which an object moves backward after an event like firing a gun. Following the Law of Conservation of Momentum, it can be calculated given no external forces act on the system. In simpler terms, its value ensures the momentum before and after the gunshot remains equal, reflecting the law that the total momentum in an isolated system is constant.
When a gun is fired horizontally, momentum from the bullet's motion is imparted to the shooter in the opposite direction. This is essential to ensure that the total backward recoil equals the momentum carried forward by the bullet.
The mathematical expression is derived from the equation:

• Initial momentum = 0 (system at rest)
• Final momentum = momentum of bullet + momentum of shooter
• Equation: \( m_b \times v_b + m_h \times v_{hx} = 0 \)

This calculates the recoil velocity \( v_{hx} \) as \( v_{hx} = -\frac{m_b \times v_b}{m_h} \), indicating a "backward" direction by the negative sign.

Horizontal Firing

Horizontal firing refers to shooting a projectile parallel to the ground. In physics problems like the one here, horizontal firing allows us to consider only the horizontal component of the forces in action without any influence from vertical forces like gravity immediately. Firing horizontally means both the bullet and hunter's gun start on the same line of action, which simplifies the analysis.
This scenario was essential for calculating the hunter's recoil velocity as only the horizontal motion needed consideration.
For horizontal firing, the bullet's complete velocity impacts the system's recoil, captured by the formula: \( v_{hx} = -\frac{m_b \times v_b}{m_h} \). The refusal of vertical components allows for a more direct application of the conservation of momentum principle.

Angle of Projection

The angle of projection is crucial in physics when discussing projectiles. It determines how a projectile splits its velocity into vertical and horizontal components. Knowing these components is key to evaluating the impact of firing projectiles at various angles.
In the given problem, the bullet is fired at an angle of \( 56.0^{\circ} \) above the horizontal. This alters the recoil because the bullet's velocity must be split into horizontal and vertical parts, where only the horizontal component affects the recoil.
Mathematically, the horizontal component is obtained using trigonometry as \( v_b \times \cos(56.0^{\circ}) \). This component then fits into our recoil equation: \( m_b \times v_b \times \cos(\theta) + m_h \times v_{hr} = 0 \). This highlights how a projectile's launch angle can drastically affect resulting calculations.

Physics Problem Solving

Physics problem-solving often involves breaking down complex scenarios into manageable parts, like in this exercise. Using conservation principles, setting clear initial conditions, and knowing which formulas to apply are critical steps in arriving at the solution.
It's important to begin by identifying known variables, like masses and velocities, which help set up the equations needed for solutions. Another essential strategy is clarifying which aspects of a problem depend on conservation laws like momentum and understanding their implication in isolated systems.
In this problem, knowing how bullets fired at angles behave differently than those fired horizontally introduced an additional layer of complexity. Each problem visualized becomes a puzzle, which we piece together using physics laws ensuring an accurate approach to understanding motion dynamics.