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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 \(56.0^{\circ}\) above the horizontal.

Short Answer

Expert verified
(a) 0.056 m/s horizontally; (b) 0.056 m/s at an angle of 56 degrees above the horizontal.

Step by step solution

01

Understand the Concept of Conservation of Momentum

Momentum is conserved in an isolated system. If a hunter fires a bullet, the total momentum before firing (which is zero) will equal the total momentum after firing. Thus, the momentum of the bullet and the hunter after firing must sum to zero.
02

Establish the Equation for Conservation of Momentum

Using conservation of momentum, the momentum of the bullet must be equal in magnitude and opposite in direction to the momentum of the hunter. The equation for conservation of momentum for part (a) is: \[ m_b \cdot v_b = - m_h \cdot v_h \]where \(m_b = 4.20 \times 10^{-3}\ kg \) (mass of the bullet), \(v_b = 965 \ m/s \) (velocity of the bullet), \(m_h = 72.5 \ kg \) (mass of the hunter), and \(v_h\) is the recoil velocity of the hunter.
03

Solve for the Horizontal Recoil Velocity of the Hunter

Substitute the given values in the equation: \[ (4.20 \times 10^{-3} \ kg) \times (965 \ m/s) = (72.5 \ kg) \times v_h \]This simplifies to:\[ v_h = \frac{(4.20 \times 10^{-3} \ kg) \times (965 \ m/s)}{72.5 \ kg} \]Solving this gives the recoil velocity \(v_h \approx 0.0559 \ m/s\). Thus, the hunter moves backward at approximately 0.056 m/s horizontally.
04

Determine the Components of Bullet's Momentum for Angled Shot

When the bullet is fired at an angle, decompose its velocity into horizontal and vertical components:\[ v_{bx} = v_b \cos(56.0^\circ) \]\[ v_{by} = v_b \sin(56.0^\circ) \]These components are used to find the recoil velocity for the angled shot.
05

Calculate Horizontal Component of the Hunter's Recoil Velocity

Apply conservation of momentum horizontally with the bullet's x-component.\[ m_b \cdot v_{bx} = -m_h \cdot v_{hx} \]Substitute the values:\[ v_{hx} = \frac{(4.20 \times 10^{-3} \ kg) \cdot (965 \cos(56.0^\circ) \ m/s)}{72.5 \ kg} \approx 0.031 \ m/s \]
06

Calculate Vertical Component of the Hunter's Recoil Velocity

The conservation of momentum in the vertical direction informs us:\[ m_b \cdot v_{by} = -m_h \cdot v_{hy} \]Thus:\[ v_{hy} = \frac{(4.20 \times 10^{-3} \ kg) \cdot (965 \sin(56.0^\circ) \ m/s)}{72.5 \ kg} \approx 0.047 \ m/s \]
07

Combine the Components to Find Total Recoil Velocity Magnitude

The resultant recoil velocity \( v_h \) is found using:\[ v_h = \sqrt{v_{hx}^2 + v_{hy}^2} \]Substitute the values to find:\[ v_h = \sqrt{(0.031 \ m/s)^2 + (0.047 \ m/s)^2} \approx 0.056 \ m/s \]
08

Calculate the Angle of Recoil

The angle \( \theta \) above the horizontal for the recoil can be found using:\[ \theta = \tan^{-1}\left(\frac{v_{hy}}{v_{hx}}\right) \]\[ \theta \approx \tan^{-1}\left(\frac{0.047 \ m/s}{0.031 \ m/s}\right) \approx 56.0^\circ \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Physics Problems
Physics problems like the example of a hunter shooting a rifle on an icy pond can offer insights into fundamental principles. Often, these problems involve breaking down complex physical interactions into simpler parts to make them easier to solve. This particular problem used the concept of the conservation of momentum to find the recoil velocity of the hunter.
In scenarios with limited external forces, like an icy frictionless surface, the law of conservation of momentum becomes really handy. This law states that the total momentum of an isolated system remains constant if no external forces act on it.
  • The system here includes the hunter, rifle, and the bullet.
  • Before the gun is fired, the total momentum is zero because the hunter and bullet are at rest relative to each other.
  • After firing, the bullet moves in one direction and the hunter recoils in the opposite direction.

Understanding such types of physics problems helps learners appreciate intrinsic properties like momentum and how they govern everyday actions. The conservation principle simplifies complex real-world interactions to manageable problems that can be calculated with straightforward equations.
Projectile Motion
Projectile motion involves two-dimensional motion under the influence of gravity. In the case of the hunter's gunshot, when the rifle is fired at an angle, the bullet path takes the form of a projectile. This introduces a vertical component into the motion of the bullet.
To solve projectile problems, one must decompose this motion into horizontal and vertical components. The horizontal component is responsible for the constant velocity of the object, while the vertical component is influenced by gravity.
  • Horizontal motion: characterized by constant velocity, since no horizontal acceleration acts on the bullet.
  • Vertical motion: influenced by gravity, meaning the bullet accelerates downwards at 9.8 m/s².
  • The bullet's velocity can be broken down as: horizontal velocity component, \( v_{bx} = v_b \cos(\theta) \), and vertical velocity component, \( v_{by} = v_b \sin(\theta) \).

For the problem at hand, understanding that even if a bullet is fired at an angle, the overall motion still needs to adhere to momentum conservation laws is crucial for finding the accompanying recoil effect. This approach gives students an understanding of how different forces and angles play a role and allow for precise calculations of motion characteristics in projectile scenarios.
Recoil Velocity
Recoil velocity is the speed at which an object moves backward after a forward action, like firing a rifle. In physics, it is an outcome of conservation of momentum. When the rifle fires a bullet, the system's total momentum—initially zero—must remain zero.
Applying this law here involves equating the momentum of the bullet to the momentum of the hunter moving in the opposite direction:
  • Calculate the bullet's momentum: \( m_b \cdot v_b \).
  • Determine the hunter's recoil velocity using: \( m_h \cdot v_h = - (m_b \cdot v_b) \).
  • This equation lets us solve for \( v_h \), the recoil velocity of the hunter.

When the bullet is fired horizontally, the recoil velocity is straightforward to calculate. However, when fired at an angle, it's necessary to consider both horizontal and vertical components of the bullet's velocity. By solving this, one can decompose the recoil motion into horizontal and vertical components and then combine them to find the resultant.
Recoil velocity provides a practical example of how conservation laws are not just theoretical notions, but actively shape interactions in physical systems, offering a bridge between mathematical formulations and real-world actions.

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Most popular questions from this chapter

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