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Chapter 8: Problem 68

A \(1000 .-\mathrm{kg}\) cannon shoots a \(30.0-\mathrm{kg}\) shell at an angle of \(25.0^{\circ}\) above the horizontal and a speed of \(500 . \mathrm{m} / \mathrm{s}\). What is the recoil velocity of the cannon?

Short Answer

Expert verified
Answer: 1.50 m/s

Step by step solution

01

Find the initial momentum of the system

Before firing the shell, both the cannon and the shell are stationary, so their initial momenta are zero.
02

Find the final momentum of the shell

To find the final momentum of the shell, first find its velocity components in the horizontal and vertical directions using the angle and speed of the projectile. The horizontal and vertical components of the shell's velocity are given by: Horizontal component: \(v_{x_{shell}} = v_{shell} \cos(25.0^{\circ})\) Vertical component: \(v_{y_{shell}} = v_{shell}\sin(25.0^{\circ})\) Substitute the given values for \(v_{shell}\) and angle to find \(v_{x_{shell}}\) and \(v_{y_{shell}}\). Now find the momentum of the shell in each direction by multiplying each velocity component by the mass of the shell: Momentum in x-direction: \(p_{x_{shell}} = m_{shell} v_{x_{shell}}\) Momentum in y-direction: \(p_{y_{shell}} = m_{shell} v_{y_{shell}}\)
03

Find the final momentum of the cannon

Since the total momentum is conserved in both horizontal and vertical directions, the final momentum of the cannon will be equal in magnitude but opposite in direction to the final momentum of the shell. Momentum in x-direction: \(p_{x_{cannon}} = -p_{x_{shell}}\) Momentum in y-direction: \(p_{y_{cannon}} = -p_{y_{shell}}\)
04

Calculate the recoil velocity of the cannon

Now we have the momentum of the cannon in both the x and y directions. To find the recoil velocity of the cannon, we can divide each momentum component by the mass of the cannon: Recoil velocity in x-direction: \(v_{x_{cannon}} = \frac{p_{x_{cannon}}}{m_{cannon}}\) Recoil velocity in y-direction: \(v_{y_{cannon}} = \frac{p_{y_{cannon}}}{m_{cannon}}\) Then, we can find the magnitude of the recoil velocity using the Pythagorean theorem: Recoil velocity: \(v_{recoil} = \sqrt{v_{x_{cannon}}^2 + v_{y_{cannon}}^2}\) Calculate and answer with the recoil velocity of the cannon.

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Key Concepts

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

Projectile Motion
Projectile motion is a two-dimensional motion where an object is thrown near the Earth's surface, and it moves along a curved path under the action of gravity. To analyze this motion, it is typical to break it down into horizontal and vertical components.
  • Horizontal Motion: The horizontal component is consistent since no forces (like air resistance, in ideal cases) are acting in this direction. This is calculated using the formula: \( v_{x} = v_0 \cos(\theta) \) where \( v_0 \) is the initial velocity and \( \theta \) is the angle above the horizontal.
  • Vertical Motion: The vertical component is influenced by gravity, meaning it accelerates downward at a rate of \( 9.8 \text{ m/s}^2 \). The initial vertical velocity is calculated by \( v_{y} = v_0 \sin(\theta) \).
Understanding these components helps predict and calculate various outcomes like range, time of flight, and maximum height in projectile problems.
Once broken down, both component calculations can be individually pursued using formulas of motion, enabling the determination of the projectile鈥檚 behavior in two dimensions.
Recoil Velocity
Recoil velocity is the speed at which an object, such as a cannon, moves in the opposite direction after ejecting a projectile, like a cannonball. This phenomenon is a perfect example of the conservation of momentum.
  • Conservation of Momentum: In isolated systems, momentum before and after an event must be equal. Hence, when a cannon fires a shell, the forward momentum of the shell is equal and opposite to the backward momentum of the cannon.
  • Recoil Calculation: Following the shot, since the initial momentum was zero (both are stationary), the total final momentum is also zero: \( p_{x_{cannon}} + p_{x_{shell}} = 0 \). Thus, \( p_{x_{cannon}} = - p_{x_{shell}} \).
By dividing this momentum by the mass of the cannon, the recoil velocities along each axis can be calculated and then combined using the Pythagorean theorem to get the magnitude of the recoil velocity.
Physics Problem Solving
Physics problem-solving requires a structured approach to ensure that complex concepts are broken down logically and are easier to tackle.
  • Understand the Problem: Begin by identifying and understanding all given information and required outcomes. Sketching diagrams can help visualize the scenario.
  • Break Down the Problem: Dissect the problem into smaller parts by identifying variables, writing down known and unknown quantities, and formulating the equations needed.
  • Apply Physics Concepts: Use relevant physics principles and equations, such as conservation laws or kinematic equations, to frame your problem.
  • Calculate and Analyze: Perform calculations carefully, ensuring units are consistent, then interpret results to check if they are reasonable. Re-evaluate steps if results are unexpected.
Following this step-by-step guidance not only helps in solving specific problems like finding recoil velocity but also strengthens your overall understanding and application skills in physics.

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Most popular questions from this chapter

When a bismuth-208 nucleus at rest decays, thallium-204 is produced, along with an alpha particle (helium- 4 nucleus). The mass numbers of bismuth-208, thallium- 204 , and helium- 4 are 208,204 , and 4 , respectively. (The mass number represents the total number of protons and neutrons in the nucleus.) The kinetic energy of the thallium nucleus is a) equal to that of the alpha particle. b) less than that of the alpha particle. c) greater than that of the alpha particle.

A catapult on a level field tosses a 3 -kg stone a horizontal distance of \(100 \mathrm{~m}\). A second 3 -kg stone tossed in an identical fashion breaks apart in the air into 2 pieces, one with a mass of \(1 \mathrm{~kg}\) and one with a mass of \(2 \mathrm{~kg} .\) Both of the pieces hit the ground at the same time. If the 1 -kg piece lands a distance of \(180 \mathrm{~m}\) away from the catapult, how far away from the catapult does the 2 -kg piece land? Ignore air resistance. a) \(20 \mathrm{~m}\) c) \(100 \mathrm{~m}\) e) \(180 \mathrm{~m}\) b) \(60 \mathrm{~m}\) d) \(120 \mathrm{~m}\)

An artillery shell is moving on a parabolic trajectory when it explodes in midair. The shell shatters into a very large number of fragments. Which of the following statements is true (select all that apply)? a) The force of the explosion will increase the momentum of the system of fragments, and so the momentum of the shell is not conserved during the explosion. b) The force of the explosion is an internal force and thus cannot alter the total momentum of the system. c) The center of mass of the system of fragments will continue to move on the initial parabolic trajectory until the last fragment touches the ground. d) The center of mass of the system of fragments will continue to move on the initial parabolic trajectory until the first fragment touches the ground. e) The center of mass of the system of fragments will have a trajectory that depends on the number of fragments and their velocities right after the explosion.

The center of mass of an irregular rigid object is always located a) at the geometrical center of c) both of the above the object. d) none of the above b) somewhere within the object.

A baseball player uses a bat with mass \(m_{\text {bat }}\) to hit a ball with mass \(m_{\text {ball }}\). Right before he hits the ball, the bat's initial velocity is \(35.0 \mathrm{~m} / \mathrm{s}\), and the ball's initial velocity is \(-30.0 \mathrm{~m} / \mathrm{s}\) (the positive direction is along the positive \(x\) -axis). The bat and ball undergo a one-dimensional elastic collision. Find the speed of the ball after the collision. Assume that \(m_{\text {bat }}\) is much greater than \(m_{\text {ball }}\), so the center of mass of the two objects is essentially at the bat.
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