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Provided that the collision is inelastic, calculate the speed after the collision:Calculate the height the two objects taken from the impact up to the maximum heigh...

Question

Provided that the collision is inelastic, calculate the speed after the collision:Calculate the height the two objects taken from the impact up to the maximum height before It startsfalling to the ground_You are pushing four blocks of masses 4.00 kg, 7.00 kg; 11.00 kg, and 17.00 kg positioned next to eachother on a horizontal frictionless surface, as shown in figure below. Find the force of contact between the7.00 kg and 11.00 kg blocks when (10 pts each)you push on the Iightest block wilh a 195

Provided that the collision is inelastic, calculate the speed after the collision: Calculate the height the two objects taken from the impact up to the maximum height before It starts falling to the ground_ You are pushing four blocks of masses 4.00 kg, 7.00 kg; 11.00 kg, and 17.00 kg positioned next to each other on a horizontal frictionless surface, as shown in figure below. Find the force of contact between the 7.00 kg and 11.00 kg blocks when (10 pts each) you push on the Iightest block wilh a 195 N force t0 the left you push on the heaviest block with a 195 N force t0 the right



Answers

Two blocks of masses $m_{1}=2.00 {kg}$ and $m_{2}=4.00 {kg}$ are each released from rest at a height of $h=5.00 {m}$ on a frictionless track, as shown in Figure ${P} 6.70,$ and undergo an elastic head- on collision. (a) Determine the velocity of each block just before the collision. (b) Determine the velocity of each block immediately after the collision. (c) Determine the maximum heights to which $m_{1}$ and $m_{2}$ rise 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.

Good day in this question we will be solving for the my stool with the given in the given collision. So we assume calculations are. And last week therefore to solver developed the cities of M one and M two. Just after the collision we will be using the equations Yeah 9 75 and 76. Where everyone I is equal to zero. So we want f is equal to do em two Divided by M one M 2. We do I and we do have is equal to M two minus M one divided by anyone that seem to be too I. So after bouncing after world, the velocity of N two becomes and do it. So after bouncing Yeah, In this there's a problem requires free one F Is equal to negative of the two F or two and two Divided by M1 last M two We too I is equal to negative and two minus M one and one divided by M one plus M two be too I. So since we do I will be canceled. This will be simplified to two M 2 you see well the negative and two minors and one Where M two it was M one if I did buy three. So we are given that M1 is equal to 6.6 kilograms. Therefore substituting We have M2 is able to 6.6 kg divided by three. So our must to see what to do. Point to you know, grabs. Thank you

For this problem, on the topic of linear momentum were shown in the figure a system of two masses. Block one has a mass one of 6.6 kg and is and is addressed on a long frictionless table against a wall. Block two has mass M two, and his place between block one and the wall. And he sent moving toward the left toward block one with a constant speed of V two. I. We want to find the value for the massive M two, for which both blocks will move with the same velocity after block to has collided once with block one and once with the wall. Now the velocities of M. One and M two just after the collision with each other are given from the conservation of momentum for elastic collisions. So V one, F is equal to two, M two of ah in one plus M two. And the final velocity of block to be to F is M two minus M one, divided by M one plus M two times V two I. Now, after bouncing of the wall, the velocity of M two becomes minus V two F. Now the problem requires the one F physical to minus V two F. So two M two over M one plus M two V two I is equal to minus M one minus M two rather minus M one, divided by M one. Bless em too, V two I. And so this simplifies to become two. M two is equal to minus M two minus M one, which means that the mass of block too, M two, that's equal to M 1/3. And so if someone is 6.6 kg, we get em too to be too 0.2 kg.


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