College Physics, Volume 1
College Physics, Volume 1
2nd Edition
ISBN: 9781133710271
Author: Giordano
Publisher: Cengage
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 3, Problem 82P

(a)

To determine

The acceleration of the bullet.

(a)

Expert Solution
Check Mark

Answer to Problem 82P

The bullet has an acceleration of 3.20×104m/s2.

Explanation of Solution

Write the kinematic equation for the motion of the bullet.

    v=v0+abΔt        (I)

Here, v is the final velocity of the bullet, v0 is the initial velocity of the bullet, ab is the acceleration of the bullet and Δt is the time taken by the bullet to leave the barrel.

Rewrite equation (I) to obtain an equation for ab.

    ab=vv0Δt

Conclusion:

Substitute 320m/s for v, 0m/s for v0 and 0.01s for Δt.

    ab=320m/s0m/s0.01s=3.20×104m/s2

Therefore, the acceleration of the bullet is 3.20×104m/s2.

(b)

To determine

The force on the bullet.

(b)

Expert Solution
Check Mark

Answer to Problem 82P

The force on the bullet is 96.0N.

Explanation of Solution

Write the equation for the force on the bullet using Newton’s second law.

    Fb=mab

Here, Fb is the force on the bullet, m is the mass of the bullet and ab is the acceleration of the bullet.

Conclusion:

Substitute 3.00×103kg for m and 3.20×104m/s2 for ab.

    Fb=(3.00×103kg)(3.20×104m/s2)=96.0N

Therefore, the force on the bullet is 96.0N.

(c)

To determine

The magnitude of force exerted on the gun.

(c)

Expert Solution
Check Mark

Answer to Problem 82P

The magnitude of force exerted on the gun is 96.0N.

Explanation of Solution

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Hence write the equation for the force exerted on the gun.

    Fb=Fr

Here, Fb is the force on the bullet and Fr is the reverse force exerted by the bullet on the gun.

Conclusion:

Substitute 96.0N for Fb.

    Fr=96.0N

Therefore, the magnitude of force exerted on the gun is 96.0N. The negative sign in the answer indicates that the reaction is in the opposite direction.

(d)

To determine

The acceleration experienced by the gun

(d)

Expert Solution
Check Mark

Answer to Problem 82P

The gun experiences an acceleration of 18.5m/s2.

Explanation of Solution

Write the equation for the force on the gun using Newton’s second law.

    Fr=Mar

Here, Fb is the force on the gun, M is the mass of the gun and ar is the acceleration of the gun.

Conclusion:

Substitute 96.0N for Fr and 5.20kg for M to solve for ar.

    96.0N=(5.20kg)arar=96.0N5.20kg=18.5m/s2

Therefore, the acceleration experienced by the gun is 18.5m/s2.

(e)

To determine

The ratio of mass m and M to that of the acceleration.

(e)

Expert Solution
Check Mark

Answer to Problem 82P

The acceleration ration is the inverse mass ratio.

Explanation of Solution

Write the ratio for mass m to M and substitute the values.

    Mm=5.20kg3.00×103kg=1.73×103        (II)

Write the ratio of the acceleration and substitute.

    abar=3.20×104m/s218.5m/s2=1.73×103        (III)

Conclusion:

From equation (II) and (III), the acceleration ration is the inverse ratio of the mass.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
A 1.4 gram bullet is shot into a tree stump. It enters at a velocity of 341 m/sec and comes to rest after having penetrated the stump in a straight line. It takes 0.00032 sec to come to a stop after striking the stump. What was the force on the bullet during impact?  (answer in Newtons)
Particle physicists have identified a type of fundamental particle called a muon, which effectively behaves like a very heavy electron. Imagine a muon of mass 1.88 × 10-28 kg is observed in a particle accelerator. It has an initial speed of 3.50 × 105 m/s. It moves in a straight line, and its speed increases to 1.25 × 106 m/s in a distance of 75.0 cm. Assume that the acceleration is constant. Find the magnitude of the force exerted on the muon.
A Bullet with a mass of 5.2 grams is fired out of a rifle.  The bullet’s initial velocity is 0 m/sec.  Its muzzle velocity is 850 m/sec (in the direction the barrel is pointing).  The barrel is 60 cm long. Assume that the acceleration inside the barrel is the same for the entire length of the barrel.  What is the acceleration in m/sec2 and what is the time it takes for the bullet to travel down the barrel?  (You must solve two simultaneous equations.    vf = vi + at ; s= vit + 1/2at2)                                        Acceleration                                   time

Chapter 3 Solutions

College Physics, Volume 1

Ch. 3 - Prob. 4QCh. 3 - Prob. 5QCh. 3 - Prob. 6QCh. 3 - Prob. 7QCh. 3 - Prob. 8QCh. 3 - The lower piece of silk in Figure 3.20 is acted on...Ch. 3 - Devise a block-and-tackle arrangement that...Ch. 3 - Prob. 11QCh. 3 - Prob. 12QCh. 3 - Prob. 13QCh. 3 - Prob. 14QCh. 3 - Prob. 15QCh. 3 - Prob. 16QCh. 3 - Prob. 17QCh. 3 - Prob. 18QCh. 3 - Prob. 19QCh. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Prob. 3PCh. 3 - Prob. 4PCh. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Prob. 9PCh. 3 - Prob. 10PCh. 3 - Prob. 11PCh. 3 - Prob. 12PCh. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - Prob. 18PCh. 3 - Prob. 19PCh. 3 - Prob. 20PCh. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - A bullet is fired upward with a speed v0 from the...Ch. 3 - Prob. 24PCh. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Prob. 29PCh. 3 - Prob. 30PCh. 3 - Prob. 31PCh. 3 - Prob. 32PCh. 3 - Your friends car has broken down, and you...Ch. 3 - Prob. 34PCh. 3 - Prob. 35PCh. 3 - Prob. 36PCh. 3 - Prob. 37PCh. 3 - Prob. 38PCh. 3 - Prob. 39PCh. 3 - You are given the job of moving a refrigerator of...Ch. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - Prob. 43PCh. 3 - Prob. 44PCh. 3 - Prob. 45PCh. 3 - Prob. 46PCh. 3 - A hockey puck slides along a rough, icy surface....Ch. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - Prob. 53PCh. 3 - Prob. 54PCh. 3 - Prob. 55PCh. 3 - Prob. 56PCh. 3 - Prob. 57PCh. 3 - Prob. 58PCh. 3 - Prob. 59PCh. 3 - Prob. 60PCh. 3 - A crate of mass 55 kg is attached to one end of a...Ch. 3 - Prob. 62PCh. 3 - Prob. 63PCh. 3 - In traction. When a large bone such as the femur...Ch. 3 - Prob. 65PCh. 3 - Prob. 66PCh. 3 - Prob. 67PCh. 3 - Prob. 68PCh. 3 - Calculate the terminal speed for a pollen grain...Ch. 3 - Prob. 70PCh. 3 - Prob. 71PCh. 3 - Calculate the terminal speed for a baseball. A...Ch. 3 - Prob. 73PCh. 3 - Prob. 74PCh. 3 - Prob. 75PCh. 3 - Prob. 76PCh. 3 - Prob. 77PCh. 3 - Prob. 78PCh. 3 - Prob. 79PCh. 3 - Prob. 80PCh. 3 - Prob. 81PCh. 3 - Prob. 82PCh. 3 - Prob. 83PCh. 3 - Prob. 84PCh. 3 - Prob. 85PCh. 3 - An impish young lad Stands on a bridge 10 m above...Ch. 3 - Prob. 87PCh. 3 - Prob. 88PCh. 3 - Prob. 89PCh. 3 - Prob. 90PCh. 3 - Prob. 91PCh. 3 - Prob. 92PCh. 3 - Prob. 93PCh. 3 - Prob. 94PCh. 3 - Prob. 95PCh. 3 - Prob. 96PCh. 3 - Prob. 97PCh. 3 - Prob. 98PCh. 3 - Prob. 99P
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Newton's Second Law of Motion: F = ma; Author: Professor Dave explains;https://www.youtube.com/watch?v=xzA6IBWUEDE;License: Standard YouTube License, CC-BY