Question

A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20-g bullets at 965 m/s. The mass of the hunter (including his gun) is 72.5 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.

Short answer

(a) -0.0560 m/s, (b) -0.0314 m/s.

Step-by-step solution

Understand the Law of Conservation of Momentum

Momentum is conserved in an isolated system, which means that the initial momentum before firing the rifle is equal to the final momentum after firing the rifle. Initially, the system is at rest, thus the total initial momentum is zero.

Express Conservation of Momentum Mathematically

The conservation of momentum can be expressed using the formula: \[ m_b \cdot v_b + m_h \cdot v_h = 0 \]where:- \( m_b \) is the mass of the bullet, 4.20 g (or 0.0042 kg).- \( v_b \) is the velocity of the bullet, 965 m/s.- \( m_h \) is the mass of the hunter and gun, 72.5 kg.- \( v_h \) is the recoil velocity of the hunter and gun.

Solve for Recoil Velocity Horizontally

Using the conservation equation:\[ 0.0042 \times 965 + 72.5 \times v_h = 0 \]Solve for \( v_h \):\[ v_h = -\frac{0.0042 \times 965}{72.5} \approx -0.0560 \text{ m/s} \]The negative sign indicates that the hunter moves in the opposite direction of the bullet.

Analyze the Recoil at an Angle

When the bullet is fired at an angle of 56.0° above the horizontal, only the horizontal component of the bullet's momentum affects the recoil velocity of the hunter.Calculate the horizontal component of the bullet's velocity: \[ v_{bx} = 965 \cdot \cos(56.0^\circ) \]

Solve for Recoil Velocity with Angle

Substitute the horizontal component of the bullet's velocity into the conservation equation:\[ 0.0042 \times (965 \cdot \cos(56.0^\circ)) + 72.5 \times v_h = 0 \]Solve for \( v_h \):\[ v_h = -\frac{0.0042 \times 965 \cdot \cos(56.0^\circ)}{72.5} \approx -0.0314 \text{ m/s} \]The negative sign once again indicates the hunter's recoil direction is opposite to the bullet's horizontal direction.

Key concepts

Physics Problems

Physics problems are essential tools in understanding scientific concepts, such as momentum conservation. In our exercise, we explored the scenario of a hunter standing on a frictionless frozen pond.
The hunter fires a rifle, and this setting clear physics problems involve calculations of velocity and momentum.
By breaking down complex phenomena into simpler parts, physics problems help us delve into the detailed operational workings of physical laws.

• Typically, physics problems require identifying known and unknown quantities.
• They involve using formulas derived from laws or theories of nature, like the law of conservation of momentum in this case.
• Comprehensive problem-solving in physics often involves analyzing different scenarios, such as movement in a horizontal direction or at an angle.

Tackling physics problems nurtures critical thinking and provides insights on applying theoretical knowledge to real-life situations.

Recoil Velocity

Recoil velocity refers to the speed at which an object moves backward in response to an action, like a gun firing a bullet. In our exercise, it's important to consider recoil velocity because it influences how entities respond when forces are applied.
Recoil velocity is evident when the hunter experiences backward momentum upon discharging a bullet forward. It's notable that the velocity is negative indicating opposite direction movement.
Recoil velocity depends on various factors:

• The mass of both the bullet and the shooter (or object experiencing recoil) is key.
• The speed at which the bullet is discharged.
• The angle at which the bullet is fired, affecting how momentum divides between horizontal and vertical components.

Understanding recoil velocity is fundamental for designing stabilizing mechanisms for firearms and ensuring the safety of operators.

Momentum Conservation Formula

The momentum conservation formula is based on the principle that the total momentum of a closed system remains constant over time if it doesn't interact with external forces.
This principle is critical in solving the exercise since it explains how the hunter's system behaves post-shot.
The formula used was: \[ m_b \cdot v_b + m_h \cdot v_h = 0 \]

• \(m_b\) is the mass of the bullet, and \(v_b\) is its velocity.
• \(m_h\) is the mass of the hunter (including the gun), and \(v_h\) denotes the recoil velocity.

The momentum conservation principle applies well in our frictionless pond scenario since no external horizontal forces are in play.
By implementing this formula, we correctly predict how changes in variables like angle of firing impact the recoil velocity. Understanding this fundamental law allows solving numerous real-world and theoretical physics problems.