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In the shown smooth ramp, 10-kg block A slides down withspeed of 3 m/s then collides with the stationary block B ofmass 9 kg. If the two blocks couple together af...

Question

In the shown smooth ramp, 10-kg block A slides down withspeed of 3 m/s then collides with the stationary block B ofmass 9 kg. If the two blocks couple together after collision,Determine the following: a) The velocity of block A just before thecollision. b) Their common velocity immediately aftercollision.

In the shown smooth ramp, 10-kg block A slides down with speed of 3 m/s then collides with the stationary block B of mass 9 kg. If the two blocks couple together after collision, Determine the following: a) The velocity of block A just before the collision. b) Their common velocity immediately after collision.



Answers

A $2 \mathrm{~kg}$ block is moving at a speed of $10 \mathrm{~m} / \mathrm{s}$ and makes a perfectly elastic collision with a second block of mass $M$, which is initially at rest. After the collision, the $2 \mathrm{~kg}$ block bounces straight back at $3 \mathrm{~m} / \mathrm{s}$. (a) Determine the mass $M$ of the second block. (b) Determine the speed of the second block after the collision.

We have two masses M&N. M. And as a number greater than equal to one, the smaller mass M is going towards the bigger mass, they're going to collide elastically and then the smaller mass big and is going to bounce off the wall in another elastic collision. And we want to answer a couple of questions about what's going to occur given certain values of And so we wanted to show, first of all that there only be one collision event is at less than equal to three. Then we want to show to collisions occur of N is equal to four. And then we want to find a number of collisions of N is equal to 10 And find the final speeds of both masses under that case of n equals 10. Because this is an elastic collision. I can use these two Formulas for the final velocity of both masses B one, F 2, F B, one, F. I can get that by taking them one minus M two over him, one plus M two times V. I one, which is denoting the initial velocity of mass one and then that's added to two times into over and one plus M two times V. I. To the initial velocity of mass to and and The final velocity of mass to is to M1 Divided by M one Plus in two times V one plus M two minus M one over N one plus into V two. So I'm gonna be able to say um Mass one and 1 is big. M mass to is gonna equal big M times in, I'm gonna say V one is Vienna and then The I two is equal to zero. And if I substitute those four things into both equations, I'm gonna arrived at V one, F is one minus N over one plus envy. Not And that V two f is gonna equal To over one plus and v not. So for there to be only one collision, this condition has to uh be satisfied. We need a negative the final velocity after the first coalition of mass one to be less than equal to the final velocity of mass to. So um what happens here is at their mass Big M collides with nm is going towards the wall at a new velocity and then it's going to reflect off the wall at the same but opposite velocity. So we going towards this direction, that negative V one F. And what we need is for that velocity to be less than equal to the final velocity and mass to because if that's true, it's always going to be trailing it. So that's why I state this inequality here. So I'm gonna, what I'm gonna do is we're gonna substitute what I have for V. One F. And V. Two F. Into that inequality. Um Go get negative one minus and over one plus in vinegar, less than equal to To over one plus and v. Not. And if you simplify this expression all the way through, you'll get that and has to be less than equal to three. So then less than equal to three. We have just one collision and part bobby, we're going to let and equal four. And we're gonna try to demonstrate that there's uh Gonna be only two collisions. So since it's greater than three, we know a second collision occurs. Now we want to develop some sort of condition in which we couldn't have a third collision. So similarly us before. So I'm gonna be um referring back to my final velocity formulas for an elastic collision. Ah This time around my initial velocity, his there's not gonna be negative VF one. And my initial velocity for the second mass is going to be the final velocity of the second mass after we had that first collision. So I'm gonna substitute the expressions that I have for each one of those into the, into both formulas so to hear as well. And if we do that, uh this is where we're going to have. So, let me call the Final Velocity. After the 2nd collision. V one, F Prime. So V one, f prime, we're gonna find that's going to equal 1-, send over one plus in negative one minus and over one plus in we're not over two and over one plus and Times to over one plus in v not. And then they velocity and mass to after the second collision would be to over one plus in negative one minus in over one plus and v not uhh Plus and -1 Over N Plus one. Um to over one plus in the not. So then similarly as before, but we need us the final velocity after the 2nd collision for the first mass V one f prime. The negative of it has to be less than equal to the final velocity Mass to after the 2nd collision. So if I plug in what I found before into the into this is a new inequality, we'll have this. Now I did a little simplification to, it'll be one minus n squared over one plus and to the second power minus four. End over one plus in to the second power less than equal to negative 21 minus and over one plus. And to the second power Plus two times in -1 Over one plus. Into the second power. If you multiply everything by one plus in squared then You arrive at 1- stands squared -4 and Less than equal to four in -4. And then if you multiply everything out and move terms Non zero turns over to the left side of the inequality you'll come up with in squared minus 10 and plus five Is less than equal to 10. Excuse me zero. And then if you're using the quadratic formula You're gonna end up, that end is equal to plus or -2 square root event. Um As a decimal, those two values will be 05-8 And 9472. So then if you draw out the real line for in The plot these two values and then use a test point for that interval. It could be any value you find that that any number that you choose him in that interval, this one over here that you're gonna satisfy the inequality That V one f prime is less than or equal to V two f prime. Yeah. So since n equals four, isn't that interval? And then there's no way that it could hit it a third time. Mm So that would show that they only have two collisions for when N is equal to four. All right, now we're gonna let an equal 10 and we want to figure out how many collisions are going to occur and under that situation. So since 10 is outside the interval from negative 5-8-9, then that automatically tells us that it's going to hit a third time. So in outside interval .5-8 9472 means we have dirt collision. So what I'm going to attempt to do is figure out the final velocity After the 3rd collision. So let's figure out the final velocity after the first collision. Um V. One F. Mhm. All right. Yeah. Using that N. is equal to 10. Yeah. Um This is what we're going to have here. Um V one F. Was gonna equal negative 9/11. V. Not and V two F is gonna equal to over 11. V. Not. Okay. So that's the velocity of each after the first collision. That we're gonna find the philosophy of both after the second collision. So, V one f will equal -9/11 9/11 v. Not. Um Plus To over 10. 11 To over 11. V. Not. Yeah, This is gonna give us -41 over 121 v not. And then V two F. Prime. It's gonna set up as to over 11 9/11. V not Plus 9/11 to over 11. V not That's going to give us 36 over 121. V not. So now I'm going to figure out the velocity after the third collision. So we're gonna call this V 1/2 double prime. So the way I'm getting this is that after I find V one F prime, this, I'm going into my the equation here For final velocities of elastic collision, plugging that into V one. I and I'm similarly doing the same thing to get V two. Okay. And then um The masses will be substituted inappropriately with envy and 10. So if you do all that, you can arrive in expression of -9/11 41 over 1 21. V. excuse me, I'm not exactly using the number negative 41/1 21 Vietnam using the opposite value of it. Because of the fact of the fact that it's going to be sort of chasing the bigger mouse. Uh And then that's going to include adding two times 10/11 36/1 21 V. Not. Mhm. This will eventually come out 2.26 V. Not. So that gives us the The final velocity after the 3rd collision. And then I'm gonna do a similar thing to get the Final velocity for mass to after that 3rd collision. It can be set up with this to over 11 41/1 21 v. Not. Mhm. Uh plus 9/11 36/1 21 B. This will eventually be .31 V. Not. So, that's the final velocity after their collision of mass to after that dirt collision. Uh Mass Big Am is going at a slower rate than or traveling as a smaller speed and then a bigger mass. And they're both going the same direction. So you're just gonna keep chasing it forever and ever. So you won't go beyond a third collision. So we would be able to say that we only have three collisions when N is equal to 10.

In this exercise, we have ah, block A that has a mass M and an initial speed V and two blocks B and C B has a mass to him and CMS em and they are initially at rest and block a collides with block B And then block V collides with Bloch si and we have to find what are the final speeds of A, B and C. So we have to consider that the three collisions, Actually, the two collisions that happened are elastic. So let's find out, um, what the speeds are by using both conservation of energy and conservation of momentum. So the first collision is the collusion of A with B. So the initial, uh, energy of a is M. V squared over two and my concern conservation of interview This is equal to the final energy of A, which is M times V. A square. Over to V is the the speed of the block a plus the mass of the block to which is to him times the speed on the block B. Which is we be. I'm gonna call VB over to So this leads two. If we just cancel the twos out and so And the the EMS as well. This leads to V squared is equal to the ace Where close to V V squared. Okay, so this is one of the equations and then we have this concern, conservation of energy, and then we have to use conservation of momentum. The initial momentum off a is M. V. The final momentum is M v A. I'm assuming that va is moving to the right. But that is not the problem. It's not, uh it doesn't restrict our options in the end, because if V is actually pointing to the left, then we're gonna get a minus sign. And the final momentum of B is MPP. So from here we have the V is equal to V A plus VB. I'm gonna isolate V A. So I have the three a is equal to the miners. VB So this is the second equation we have and I'm gonna just love this into the first equation. So the first vision is the square is able to be a squared and these female sgb plus two times vb I swear So the square is equal to the squared minus two v vb square VM Sorry, because we be square plus two VB square. So is that the V's Cancel out each and we get to, Ah, one of the babies do as well So have to be. It's about a three d d. So we get that VB is equal to two times V over V over three insert. This is the speed of the block V after, uh, after Block A heads it and now confined the speed of the block. A. So that's just V amenity bees, which V minus two Viewer three That's minus V or three notice. That's since there's a minus sign here than a The Block A is moving to the left, and it will not collide with the other blocks anymore. So this year is the final speed a block A. Okay, uh, now we can move on to the second collision between about being black sea, so it's B B with C. Again, we have to use conservation of energy. So we have the initial speed, which is just outside of the initial energy, which is M V C B squared over two, and this is equal to ah, the final area that's in actually the massive look these to him, so I should everything to him here. So it's the method. Luck be times the final speed of B square. That's capital T B divided by two plus the massive block C, which is just M times the finals beat of C squared over two. And from here we just get that too. Uh, the M skin. Salo's so does too. So we get the to be be square is equal to capital VB Square. Actually, there's two years to capital B B squared plus B C square, and I'm gonna save this equation and now use the conservation of momentum. So from consideration when that's when we get a to M V B is equal to two m sir, I should add that the initial momentum. Miss, just lower case V B, This is able to to em capital Phoebe plus M b c. So again, the EMS cancel out and we get to the B is equal to two capital VB plus V c. And I'm gonna isolate the bees we get. The thebe is equal to lower case V B minus V C word shoe. I'm gonna take this value of the B end, Plug it into the laying in the first boots. I have to V B Square, which is able to chew times Capital VB Square. So it's this is or case vb times V C Square plus V C Square, so get to VB. Squared is equal to two V b squared, minus two times VB V C plus B C squared over two plus B C square. So the two V B squared cancels out. One of these devices do as well. So you get to VB is equal to three halves of the sea, so V C is equal to four thirds of VB. You know VB Lorca's Phoebe. Um, we have already calculated it, so it's people to to be over three. Let's just plug it in here so have four thirds times two thirds of V. So this is 8/9 of the This is the final speed of the Block C. And I can company the finest Peter the block be going back to this equation right here to have vb. VB is equal to lower case V B minus this year or two now, Phoebe, Laura case. Phoebe is just two thirds of V while the sea is eight v overnight. Then I have to multiply by two. So he gets, um, 6/9 off he That's just two thirds of the minus four V or nine. So this is equal to to view everything. This is the final speed of the block B in the end, to a question.

Our question says to look at figure nine. That's 64 in which Block A has a massive 1.6 kilograms today and Block B has a massive 2.4 kilograms and Sabi and there it's moving along a frictionless plane, the direction of the three velocity it The three velocities are gonna be appropriated as I or F for initial our final indications V A. Set by the initial velocity of la Caisse 5.5 meters per second. Vehbi survive is 2.5 meters per second for the initial velocity of block B. The final velocity of block B is given before 0.9 meters per second and it wants us to find a the speed and be the direction left or right of the velocity of block A for the final and then the collision. Whether or not it's a last acre in elastic, so toe do part a indicate this is part of a we're gonna use conservation of momentum, which the initial momentum is in a times that initial velocity of a plus instead be times the initial velocity of B is equal to the final momentum of a plus. The final moment of B. We know everything here except for the final velocity of A which is what we're asked to solve so we can arrange this equation to sulfur. The final velocity of a it's going to be in may be a I plus Well, it's gonna be M v v b I Minus in the times the final velocity of be So the massive B is common in both of those. So just to simplify this a little bit, this is gonna be n b times v b I minus bbs. And then this is all divided by the massive, eh? So, playing everything into this expression, this comes out to be 1.9 meters per second. Okay? Huh? And then for part B rass isn't moving to the left or right. But since it's positive it has to be moving to the right so we can right here. Okay. Never Part c were asked if it's an elastic Aaron elastic collision. So to do that, let's compare the initial in the final energies. So 1/2 in they times the initial velocity of a squared plus 1/2 and B times the initial velocity of B squared it Doesn't I use into this expression? This comes out to be 31.7 jewels. Okay, let's do the same thing for the final kinetic energy 1/2 M B the B f squared plus 1/2 in a the F squared. OK, pluck those values in the expression and it's also equal to 31.7 jewels. So since k initial equals K final, this is an elastic okay quicken box that in is their solution for C.


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