A .22 caliber rifle bullet traveling at 350 m / s strikes a large tree and penetrates it to a d

A .22 caliber rifle bullet traveling at 350 m / s strikes a...

A .22 caliber rifle bullet traveling at $350 \mathrm{~m} / \mathrm{s}$ strikes a large tree and penetrates it to a depth of $0.130 \mathrm{~m}$. The mass of the bullet is $1.80 \mathrm{~g}$. Assume a constant retarding force. (a) How much time is required for the bullet to stop? (b) What force, in newtons, does the tree exert on the bullet?

Key Concepts

Newton’s Second Law of Motion

Newton’s second law explains how the acceleration of an object is determined by the net force acting upon it and its mass. This fundamental principle is used to calculate the force exerted on an object by multiplying its mass with its acceleration, which in this case helps determine the retarding force applied to stop the bullet.

Uniform Acceleration Kinematics

This concept involves the equations of motion for an object experiencing constant acceleration or deceleration. It provides the tools needed to relate initial velocity, displacement, time, and acceleration, allowing one to calculate the duration required for an object to come to a stop when decelerating over a known distance.

Transcript

00:01 All right, so we have a rifle firing a bolt into a tree problem, and then the bullet comes to a halt.

00:10 However, the bullet was originally traveling at 350 meters per second.

00:15 It's, you know, again, coming to a halt.

00:17 It's final velocity to zero.

00:19 Within the bark of the tree, it traveled 13 centimeters.

00:24 So we're going to call that 0 .13 meters.

00:27 And then the mass of the bullets is 0 .18 grams, or sorry, 1 .8 grams, but we're going to go ahead and convert that to kilograms right away because we want everything to be in si units.

00:40 So part a is asking us, what time interval or how long did it take to do this? and so because we don't know what the acceleration is, we're going to use the kinematics equation that ignores acceleration.

00:53 So that's finding the average velocity, and then multiplying it by the time interval.

00:59 And then we're going to want to rearrange that to solve for the time interval because that's what we're interested in, and then quickly get rid of that final velocity because we just discussed that it is equal to zero.

01:11 So the time interval is going to be equal to 2 delta x over v .0, and that is going to result in something, but we have to plug in the numbers first.

01:25 So 0 .310, 0 .130 for displacement, 350 for velocity, and then we get that it was 7 .43 times 10 to the minus 4 seconds to three significant figures.

01:43 So very, very quick interaction.

01:47 Whoops, sorry...

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