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Problem 50 Medium Difficulty

A 1200 -kg station wagon is moving along a straight highway at 12.0 $\mathrm{m} / \mathrm{s}$ . Another car, with mass 1800 $\mathrm{kg}$ and speed 20.0 $\mathrm{m} / \mathrm{s}$ , has its center of mass 40.0 $\mathrm{m}$ ahead of the center of mass of the station wagon (Fig. 8.39$) .$ (a) Find the position of the center of mass of the system consisting of the two automobiles. (b) Find the magnitude of the total momentum of the system from the given data. (o) Find the speed of the center of mass of the system. (d) Find the total momentum of the system, using the speed of the center of mass. Compare your result with that of part (b).

Answer

(a) $x_{c m}=24.0 \mathrm{m}$
(b) $p_{\text { tot }}=50400 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$
(c) $v_{c m}=16.8 \mathrm{m} / \mathrm{s}$
(d) $p_{\text { tot }}=50400 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$

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Video Transcript

we know that the position vector off the center off Mars off the system off particles is given by the situation equation number one. Well, I indicates to the ice particle. We also know that the total momentum off a system off of particles equals it's total mass multiplied by the velocity off the center off mass. So we can write the second equation According to that and where the moment, um, off each individual particle is given by P I is a culture m I t. Now in the first case taking the origin at the center, off mass off the yellow card and plus X access directed towards the center off mass toe the rain card. So the position off each center off mass is X y. Is it called a zero and Excel is a call. Judy is a call to 40 m. I hope we have only X component. So applying equation going to the system off the two cars so we will get X C m is equal to m by x y plus m r x r upon mm bye plus m r. So now we plug our values for M by M. R. we why we are so we get Xia physical too. 1200 in 20 plus 1800 plus into 40 upon 1200 plus 1800 so we will help eggs. C M is a culture 24 major twin center off Mass. So the center off mass off the system off the two cars plays between the center off mass off the yellow car and the center off Mass off the red card, which is 24 m from the center off Mass off the yellow car. Now, in the next case, we plug our values for em by and Viv I into Equation three. That means in this situation the turban. So we relegate B Y is equal to 1200 into 12 which is equivalent to 14400 kilogram into me. The 1st, 2nd and flubber values for Emma and we're into Equation three. So we'll get the mo mentum off R B R is equal to 1802 20 equivalent to 36,000 kilogram into meter per second. So the total mo mentum off the system off the two cars is be total is a call to B Y plus B r and the some off this. 14,400 36,000. We'll give you the answer 50,400 kilograms into meet up on second Now reading the whole system as a particle with total Mars and musical toe Mbai plus m r. We can get the velocity off the system by plugging our values toe be total and AM into equation three Sylvie gate P total is equal tomb 1200 plus 1800 into the S. Yeah, so from this we will get V s is equal tomb 16.8 m per second. The velocity of the system is 16.8 m per second, which is the velocity off its center off months. So we can write BCM is a call to 16.8 m per second. Now we plug our values off M and V C M into equation too. So the gate the total is a pool too 1200 plus 1800 into 16.8, which gives the answer. 50,400 kilogram me joke was second, which is the same result as part me mhm