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At a classic auto show, a $840-\mathrm{kg}$ 1955 Nash Metropolitan motors by at 9.0 $\mathrm{m} / \mathrm{s}$ , followed by a $1620-\mathrm{kg} 1957$ Packard Clipper purring past at 5.0 $\mathrm{m} / \mathrm{s}$ , (a) Which car has the greater kinetic energy? What is the ratio of the kinetic energy of the Nash to that of the Packard? (b) Which car has the greater magnitude of momentum? What is the ratio of the magnitude of momentum of the Nash to that of the Packard? (c) Let $F_{N}$ be the net force required to stop the Nash in time $t$ , and let $F_{\mathrm{p}}$ be the net force required to stop the Packard in the same time. Which is larger: $F_{\mathrm{N}}$ or $F_{\mathrm{P}}$ ? What is the ratio $F_{\mathrm{N}} / F_{\mathrm{p}}$ of these two forces? (d) Now let $F_{\mathrm{N}}$ be the net force required to stop the Nash in a distance $d$ , and let $F_{\mathrm{p}}$ be the net force required to stop the Packard in the same distance. Which islarger. $F_{\mathrm{N}}$ or $F_{\mathrm{P}}$ ? What is the ratio $F_{\mathrm{N}} / F_{\mathrm{P}} ?$

(a) $\frac{K_{\mathrm{v}}}{K_{\mathrm{p}}}=1.68$(b) $\frac{p_{\mathrm{N}}}{p_{\mathrm{p}}}=0.933$(c) $\frac{F_{\mathrm{N}}}{F_{\mathrm{p}}}=\frac{p_{\mathrm{N}}}{p_{\mathrm{p}}}=0.933$(d) $\frac{F_{\mathrm{N}}}{F_{\mathrm{p}}}=\frac{K_{\mathrm{N}}}{K_{\mathrm{p}}}=1.68$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

Rutgers, The State University of New Jersey

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So in this exercise, I have two cars. Car A with mass off 840 kg and velocity off 4 m per second. And a car be that has a mass saw, Uh, 1620 kg and velocity of 5 m per second in question. A. We want to calculate the kinetic energy off car A the kinetic energy off Karbi and then calculates the ratio. Uh, que a over kb. Okay, so we know the definition of kinetic energy, So Okay, a is he could should the mass of car a times a velocity squared over two. And so we student the values we have 860 uh, times nine squared over two. This is equal Thio three. Uh, sorry. 3.4. Time send to the for JAL. So this is the first answer. Now, the second answer, K B is equal to the mess off Carvey. So 1620 times velocity squared. I squared over two, and this is equal to 2.2 times 10 to the far JAL. This is the second answer. And now the ratio Okay, a over kb. So Okay, a over K B so is equal to 3.4 times sent to the four over 2.2 times 10 to the four, and this is equal to 1.68. So this is a fraction now question be asked us to calculate the linear momentum off both cars. So why should be we want to know p a PB. And we also want to know the ratio between p A and P bee. So you want to know p a over b b? Okay, so p A. From definition off, linear momentum is equal to the mass of car eight times the velocity. So you have that It's 860 times nine, and this is equal to 7.56 times sent to the tree kilograms meter per second. Now the moment, um off Carby. So the massive carby 101,620 kilograms sounds the velocity five and this is equal to, um 8.10 times, then to the third kilograms meters per second. And now way have have to calculate P a over B b. So, um, sorry. So be a over B B. So 7.56 times 10 to the third over eight points 10 times 10 8.7 times sent to the church. And this is equal to 0.933 Hey, On question, See? Asked us to calculate what? Uh, so sorry. So questions he asked such calculate that given that it takes a well, a force s a to stop car A in a time t And given that eyes required a force f b to stop the car be in a time in the same time t so this time equal to this time. Okay, So given these forces, which frost is great and what is the ratio between the forces? So we also want to know what is f a over f be given that f a Is the force required to stop car A in a time T and F B is a force required to stop car be in the same interval off time. Okay, so we can recall that one of the nation off force is that force is equal to the time variation off the linear momentum. Okay, so given that the force is constant over time, we have that p a Is it going to f a times Delta T or just time? See, uh, and PB is equal to f b times the same adult ity. Okay, Now, given the data T in P A is equal to delta t N p b. We can compare the two forces by looking at the linear momentum. So we know that the linear momentum off car A is smaller than the linear momentum off Karbi. So by this, we have that the force that is going to be applied on car A is going to be smaller than the force that will be applied on Karbi and the racial will just be so f a over FB is going to be equal to p A over Delta T oh, divided by PB over Delta T we canceled out a tease, and we have that. This is just the racial between the linear momentum that we found in question be to be equal to 0.933 Okay, no question. See, uh, sorry question. The asks us that, given that now f a is the force required to stop the car A in a distance? The d T Delta T delta d sorry. And that if FB is the force required to stop car be in the same distance Delta D Which force would be greater now? Now we have that, uh, the work turned by the forced f a. So w a is going to be equal thio the force times the distance on which the force was applied. So distance Delta D okay, and w b will be equal to the force applied on Carvey during the distance. Delta d So from the work, Um, from the work energy theory Um, we have that this work, the EU a will have to be equal to the difference in the kinetic energy. So the initial kinetic energy Sorry. The final kinetic energy off our A minus the initial kinetic energy off car A. Okay. And, um, and given that delta T Delta D sorry is the distance it requires two for the car to stop. So the final kinetic energy will be zero, because the car will be stationary and we only have minus the kinetic energy off car A, which we calculated on exercise A to be equal to two. Sorry to be equal to three point four times 10 to the three jewels. Okay. And the same for Carby. So the final minus the initial the final kinetic energy will be zero. Because is the is when the car we're evaluating at the point when the car is a stationary. So I have that will be equal to minus K B. And this is equal to from exercise a 2.2 times 10. Sorry, this is not the third sister the for to the fore. Okay, now we can, uh, see that f a. Uh, since the delta D is going to be equal for both cases, we can compare f A and F B by the kinetic energy. So we have that kind of energy on car A is greater than the kinetic energy on Karbi. So in this time, we'll have that f a will be greater than FB. Okay? And also, we can do f a over FB by passing this term here, dividing in the site and this term dividing in the side. And we have that f a over f B is equal. Thio minus 3.4 times stand to the fore over Delta D, divided by minus 2.2 times Stand to the four divided by Delta T we can cancel out the Delta tease. The minus sign also vanishes, and we have that This is just a fraction that we calculated on question A. This is just k A over KB, which we found in question each vehicle to 1.68 so this concludes exercise.

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