/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 A \(7.00-\mathrm{g}\) bullet mov... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 36

A \(7.00-\mathrm{g}\) bullet moving horizontally at \(200 \mathrm{~m} / \mathrm{s}\) strikes and passes through a 150 -g tin can sitting on a post. Just after impact, the can has a horizontal speed of \(180 \mathrm{~cm} / \mathrm{s}\). What was the bullet's speed after leaving the can?

Short Answer

Expert verified
The bullet's speed after leaving the can is approximately 161.43 m/s.

Step by step solution

01

Convert Units

First, we need to convert the speed of the can from centimeters per second to meters per second for consistency in units. The can's speed is given as 180 cm/s. Thus, the conversion is:\[180 \text{ cm/s} = 1.80 \text{ m/s}\]
02

Identify Known Values

Next, list what we know:- Initial mass of the bullet, \(m_b = 7.00 \mathrm{~g} = 0.007 \mathrm{~kg}\)- Initial speed of the bullet, \(v_{b_i} = 200 \mathrm{~m/s}\)- Mass of the can, \(m_c = 150 \mathrm{~g} = 0.150 \mathrm{~kg}\)- Final speed of the can, \(v_{c_f} = 1.80 \mathrm{~m/s}\)
03

Apply Conservation of Momentum

Conservation of momentum states that the total momentum before the collision is equal to the total momentum after. For this collision:Initial momentum: \(m_b v_{b_i}\)Final momentum: \(m_b v_{b_f} + m_c v_{c_f}\)Setting these equal, we have:\[m_b v_{b_i} = m_b v_{b_f} + m_c v_{c_f}\]
04

Solve for Bullet's Final Speed

Rearrange the conservation of momentum equation to solve for the bullet鈥檚 final speed \(v_{b_f}\):\[v_{b_f} = \frac{m_b v_{b_i} - m_c v_{c_f}}{m_b}\]Substitute the known values:\[v_{b_f} = \frac{(0.007)(200) - (0.150)(1.80)}{0.007}\]Calculate to find \(v_{b_f}\).
05

Calculate Result

Perform the calculation:\[v_{b_f} = \frac{1.4 - 0.27}{0.007} = \frac{1.13}{0.007} \approx 161.43 \mathrm{~m/s}\]The bullet's speed after leaving the can is approximately 161.43 m/s.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Collision Physics
Collision physics deals with the interaction between two objects that come into contact, exchanging forces and energy. In our original exercise case, we have an example of an inelastic collision: the bullet striking and passing through the tin can. Inelastic collisions typically involve a decrease in kinetic energy. Unlike elastic collisions, energy might not be conserved as some of it transforms into sound, heat, or deformation energy. However, the total momentum is always conserved, which brings us to the principle of momentum conservation.

Important points include:
  • The nature of the collision (elastic or inelastic).
  • Conservation of momentum applies in all collision types.
  • Energy conservation might not always hold as energy could be lost to deformation or other forms.
Understanding the type of collision helps predict the motion and final state of the objects involved. In this instance, although the bullet changes speed and kinetic energy, the momentum principle helps us solve the exercise.
Momentum Transfer
Momentum transfer refers to the process of momentum being passed from one object to another during a collision. In the scenario of the bullet and the tin can, the bullet's momentum is transferred to the can, causing the tin can to move while the bullet continues with a reduced speed.

Mathematically, momentum transfer during a collision can be expressed as:
  • Initial momentum: Sum of momenta of both objects before interaction.
  • Final momentum: Sum of momenta of both objects after interaction.
  • The equation: \[m_b v_{b_i} = m_b v_{b_f} + m_c v_{c_f}\] where \(m_b\) and \(m_c\) are the masses, and \(v_{b_i}, v_{b_f},\) and \(v_{c_f}\) are the initial and final velocities.

The principle of conservation dictates that the total momentum before and after the collision remains constant. This helps us predict the post-collision speeds of the objects.
Unit Conversion in Physics
Unit conversion is an essential skill in physics to ensure consistency and accuracy in calculations. In the context of the original exercise, converting the velocity of the tin can from centimeters per second to meters per second is a crucial first step.

Standard conversions often used include:
  • 1 meter = 100 centimeters.
  • Velocity in cm/s converted to m/s (divide by 100).

To convert 180 cm/s to m/s, divide by 100, yielding 1.8 m/s. This ensures all quantities are in the correct units before applying any formulas.

Proper unit conversion minimizes errors and ensures clarity in understanding, especially in problem-solving scenarios that require interactions between objects, like our bullet and can setup.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two balls of equal mass approach the coordinate origin, one moving downward along the \(y\) -axis at \(2.00 \mathrm{~m} / \mathrm{s}\) and the other moving to the right along the \(-x\) -axis at \(3.00 \mathrm{~m} / \mathrm{s}\). After they collide, one ball moves out to the right along the \(+x\) -axis at \(1.20 \mathrm{~m} / \mathrm{s}\). Find the scalar \(x\) and \(y\) velocity components of the other ball. This is a two-dimensional collision and momentum must be conserved independently in each perpendicular direction, \(x\) and \(y\). Take \(u p\) and to the right as positive. Accordingly, keeping in mind that before impact only one ball had an \(x\) -component of velocity, Or $$ \begin{aligned} \text { (Momentum before) }_{x} &=\text { (Momentum after) }_{x} \\ m(3.00 \mathrm{~m} / \mathrm{s})+0 &=m(1.20 \mathrm{~m} / \mathrm{s})+m v_{x} \end{aligned} $$ Here \(v_{\mathrm{x}}\) is the unknown \(x\) -component of velocity of the second ball acquired on impact. Since we know that the first ball lost some of its \(x\) -momentum, the second ball must have gained it. Moreover, (Momentum before) \(_{y}=(\text { Momentum after })_{y}\) or $$ 0+m(-2.00 \mathrm{~m} / \mathrm{s})=0+m v_{y} $$ Here \(v_{\mathrm{y}}\) is the \(y\) -component of velocity of the second ball. (Why the minus sign?) Solving each equation, after cancelling the mass we find that \(v_{x}=1.80 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=-2.00 \mathrm{~m} / \mathrm{s}\)

(a) What minimum thrust must the engines of a \(2.0 \times 10^{5} \mathrm{~kg}\) rocket have if the rocket is to be able to slowly rise from the Earth when aimed straight upward? \((b)\) If the engines eject gas at the rate of \(20 \mathrm{~kg} / \mathrm{s}\), how fast must the gaseous exhaust be moving as it leaves the engines? Neglect the small change in the mass of the rocket due to the ejected fuel. [Hint: Study Problem 8.20.]

Three masses are placed on the \(y\) -axis: \(2 \mathrm{~kg}\) at \(y=300 \mathrm{~cm}, 6 \mathrm{~kg}\) at \(y=150 \mathrm{~cm}\), and \(4 \mathrm{~kg}\) at \(y=-75 \mathrm{~cm}\). Find their center of mass.

What force is exerted on a stationary flat plate held perpendicular to a jet of water as shown in Fig. 8-5? The horizontal speed of the water is \(80 \mathrm{~cm} / \mathrm{s}\), and \(30 \mathrm{~mL}\) of the water hit the plate each second. Assume the water moves parallel to the plate after striking it. One milliliter (mL) of water has a mass of \(1.00 \mathrm{~g}\). This question deals with speed, mass, time, and force, and that suggests impulse-momentum and Newton's Second Law. The plate exerts an impulse on the water and changes its horizontal momentum. The water exerts a counterforce on the plate. Taking the direction to the right as positive, \((\text { Impulse })_{x}=\) Change in \(x\) -directed momentum $$ F_{x} \Delta t=\left(m v_{x}\right)_{\text {final }}-\left(m v_{x}\right)_{\text {initial }} $$ Let \(t\) be \(1.00 \mathrm{~s}\) so that \(m\) will be the mass that strikes in \(1.00 \mathrm{~s}\), namely \(30 \mathrm{~g}\). Then the above equation becomes $$ F_{x}(1.00 \mathrm{~s})=(0.030 \mathrm{~kg})(0 \mathrm{~m} / \mathrm{s})-(0.030 \mathrm{~kg})(0.80 \mathrm{~m} / \mathrm{s}) $$ from which \(F_{x}=-0.024 \mathrm{~N}\). This is the force exerted by the plate on the water. The law of action and reaction tells us that the jet exerts an equal but opposite force on the plate.

A ball of mass \(m\) at rest at the coordinate origin explodes into three equal pieces. At some instant, one piece is on the \(x\) -axis at \(x=40 \mathrm{~cm}\) and another is at \(x=20 \mathrm{~cm}, y=-60 \mathrm{~cm}\). Where is the third piece at that instant?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Electricity

Read Explanation

Waves Physics

Read Explanation

Measurements

Read Explanation

Energy Physics

Read Explanation

Radiation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.