A rifle with a barrel length of 60 cm fires a 10 g bullet with a horizontal speed of 400 m/s. The bullet strikes a block of wood and penetrates to a depth of 12 cm. b. How long does it take the bullet to come to rest?

Video transcript

Hi, everyone in this practice problem. We're being asked to calculate the time required to stop a round inside an armor fast. We will have a body armor fast made of polymeric composite, stopping a 15 g gone round moving horizontally at a speed of 100 and 75 m per second. After traveling a distance of five millimeter inside it, we're being asked to calculate the time required to stop the round inside the armor fast. And the options given are a three microseconds, B 27 microseconds, C 57 microseconds and D 534 microseconds. So we will model the round as a point like particle. And when entering the armor fast, the round is subjected to three different forces. First is going to be the weight uh acting vertically downward as usual. Second is going to be the normal force acting vertically upward. And lastly, it's going to be the friction force F K, which is the friction force inside the armor that is going to be stopping the round from moving. So we assume that the acceleration A X is going to be constant throughout the overall um process. And using the kinematic equation without time, we will then be able to calculate A X. And from there, we will be able to calculate the time required to stop the round inside the armor fastest. So first, let's start with um using the kinematic equation. So it is given the V I is going to be 175 m per second. The V F is going to be when the round actually stops, which is zero m per second. And the delta X is going to be five times 10 to the power of negative three m, which is coming from the five millimeter. The kinematic equation that we're going to use is going to be V F squared minus V I squared equals to two A X multiplied by delta X. So we will substitute uh all the known information here to solve for A X so that A X will then be V F squared minus V I squared. All of that divided by two delta X substituting all of our information, we have zero m per second squared for V F squared minus V I squared, which is going to be 100 and 75 m per second squared. All of that divided by two delta X which is two multiplied by five times 10 to the power of negative three m calculating this, this will give us an A X value of negative three point oh six times 10 to the power of six m per second squared. So to find the time required to stop the round, we will apply the next uh kinematic equation which is going to be V F equals to A X T plus V I. So this is uh neutralizing the acceleration that we just found to then calculate the time that it is required to stop the round inside the armor fast. So then we will substitute all of our known information to get an equation for T. So T will then equals to V F minus V I divided by A X and V F is zero m per second. V I is going to be 100 and 75 m per second. And all of that is going to be divided by A X which is going to be negative three point oh six times 10 to the power of six meters per second squared. That will essentially give us a time value of 5.7 times 10 to the power of negative five seconds, which will come out to equals to microseconds. So 57 microseconds will be the time required to stop the round inside the armor fast, which will actually correspond to option C in our answer choices. So answer C with the time required to stop the run inside the armor fast being 57 microseconds will be the answer to this practice problem and that'll be it for this video. If you guys still have any sort of confusion. Please make sure to check out our other lesson videos on similar topics and that'll be it for this one. Thank you.