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Lqure "cur larce; dcl &nl 2in loncshoan ahere Fi INR ZRhulo4ie Ihicuah pcnt 47500*0.5 s2W: M1.W $52Slcon459tNMm 32.95 Nm 063Nm405k9 bellof day moving to th...

Question

Lqure "cur larce; dcl &nl 2in loncshoan ahere Fi INR ZRhulo4ie Ihicuah pcnt 47500*0.5 s2W: M1.W $52Slcon459tNMm 32.95 Nm 063Nm405k9 bellof day moving to the ight ai 3 4 m/s collides in a head-on collis n With a 21kg ball of dyy moving to the kft at 5. 4 mys The two balls sbck together aiter the collision What is therr {d and direcbon 0l motion afler ine collision?Seedtcne5,02 mksto the leltb 5.02 ms to the right(371 MVs to the nghid. 3,71ms to the letWhich ol the following cotege concem

lqure "cur larce; dcl &nl 2in lonc shoan ahere Fi INR ZR hulo4ie Ihicuah pcnt 47 500* 0.5 s2 W: M 1.W $52 Slcon 459tNM m 32.95 Nm 063Nm 405k9 bellof day moving to the ight ai 3 4 m/s collides in a head-on collis n With a 21kg ball of dyy moving to the kft at 5. 4 mys The two balls sbck together aiter the collision What is therr {d and direcbon 0l motion afler ine collision? Seedtcne 5,02 mksto the lelt b 5.02 ms to the right (371 MVs to the nghi d. 3,71ms to the let Which ol the following cotege conceming inelastic collisions? Select one: The tota kinetic energy and the total momentun are both NOT conserved during the collision The colliding objects must stick together after the collision Only InetOu momenium consGNvco dunno the collision The total kinetic energy and the total momentum are both conserved during the collision



Answers

From conservation of linear momentum along the direction of incident ball for the system consists with colliding ball and phere $$ m v_{0}=m v^{\prime}+\frac{m}{2} v_{1} $$ where $v^{\prime}$ and $v_{1}$ are the velocities of ball and sphere 1 respectively after collision. (Remember that the collision is head on). As the collision is perfectly elastic, from the definition of co-efficeint of restitution, $$ 1=\frac{v^{\prime}-v_{1}}{0-v_{0}} \text { or, } v^{\prime}-v_{1}=-v_{0} $$ 101 Solving $(1)$ and $(2)$, we get, $v_{1}=\frac{4 v_{0}}{3}$, directed towards right. In the C.M. frame of spheres 1 and 2 (Fig.) Also, $\overrightarrow{r_{1 C}}=-\overrightarrow{r_{2 C}}$, thus $\overrightarrow{\boldsymbol{M}}=2\left[\overrightarrow{r_{1 C}} \times \widetilde{\vec{p}_{1}}\right]$ As $\overrightarrow{r_{1 C}} \perp \ddot{p_{1}}, \quad$ so, $\vec{M}=2\left[\frac{l}{2} \frac{m / 2}{2} \frac{4 v_{0}}{3} \hat{n}\right]$ (where $\hat{n}$ is the unit vector in the sense of $\overrightarrow{r_{1 C}} \times \widetilde{\vec{p}_{1}}$ ) Hence $\tilde{M}=\frac{m v_{0} l}{3}$

This question is about a collision of two spheres. He still one s and 10.5 kg And the initial velocity, It's event to be to I minus gj last k. You just put second and this against your ass mess 1 25 kg. Their initial velocity is given to be minus. I class to Jane minus cheeky you just for a second. Okay, then. That's a, um even that The final velocity off and one yes, minus I class she chain minus its k. He wants to find me to f. Okay, so to do this will be using conservation of momentum. So we get this and one b one I us em to me too. I it goes to and one you want f pass and to be two f So putting in the numbers, I'm going to use the column. Former practice tu minus tree one. This is for everyone and for him to be, too. I This is what we have. And then and one B one f plus 1.5. Me too. Okay, so, uh, rearrange. So you generate the right hand side of the left hand side. Done bringing in uh, um, turns from there. Right hand site. You can see that this throughout the course of zero, which means that you two f yes. Zero. Yeah. And then, um, you look at, um, compare kinetic energy. So Okay, candy energies before and after collision. Okay, So can the energy before is ah, half everyone and one we want I square plus have them to me to I square everything, putting in the numbers I am. One is your five and completing everyone I square will be something that squares off the components. Yes, if you repeat a step for, uh and to And then you calculate this any shaking adding energy to be 14 juice and then the final looking at the energy. Uh, it's just a and one B one f skway happened when we went ex quay. So Okay, so this is 18.5 shoes. So we have k I, uh, less than chaos. So which means that a collision in this case with me, uh, elastic. Yeah, it's not completely. It's not perfectly any lasting, but it's just last. Okay, so this is, uh, the answer for a Yeah, that's right. Stiff then for parts B. So you assume that, uh, final velocity off and one is given to be, uh, my name is your 0.25 I class to your point. Third time J minus two. Okay, he does put second. Then we wants to find me to wrap a gain. So, uh, we just we just repeat from this equation. So this is the part that needs to be changed. So we have 1.5 me to f is equal to my name is 0.5 1.5, minus hole, minus five. This is anyone and everyone else is even in such. Okay, this is found to be is your 0.375 1.1 to 5 my ministry. And then after the writing, 1.5 both sides, he gets me to have to be. Ah, what is your 0.25 I see Appointed by J and minus two. Okay. And if this one Yeah. So, uh, the collision is affect me in lasting. Okay? Because the both spheres have the same velocity final velocity after the collision. Okay, so the answer for that's B. All right, so we move on to proxy. Okay? Now, um, given that b you want f is minus. I plus d j has a k you just for a second. Uh, you want to find me to African? Okay. So to do this on the conservation of momentum. Okay, so we have, um, 1.5 me to f eyes equal to minus 51.5 or my ass for I guess, your 0.5. When it's one tree A, it is equal to 00 Uh, minus school. My name? 0.5 a, um, back there. Okay. So, me too. F you think we're too 1/1 45 minus four, plus your 40.5 e. Okay, you don't split second. Okay? Okay. Then you need another equation because the collision is, uh, you lasting. So the we have conservation off connecting energy. Okay, so that means, um huh. And one, we want every square that's half them to you to ask. Where is a photo? 14 Jews. It is the number that we are thing completed. Impress me. So putting in the numbers or, uh, finally getting energies. So So this is the final kinetic energy for M one. And then for him to Okay, So this is the equation for the conservation of kinetic energy. Ke So, uh, so continue to do the math. Um okay, so continue. Yeah. So when I do this, I'm not declining three throughout the whole equation. So this is the question I get and then simplify. You get this quite reading Christian to a squared. Plus, it's a minus. 37 goes to zero. So here you used the quadratic formula English to, uh, minus eight. Last minus, uh, be square rightness. Who I a see? Square it and then divide by four. So from here, get a go, Sue 2.74 oh, minus 6.74 Able, our values are possible. So we can completely finally velocity fall m two for each case for any goes to 2.74 you have me to have to be Ah, minus 3.58 Okay, you just go second. And then for a he goes to minus 6.7 full. You have you to half to be minus your point for one night. Okay? You just put thinking okay, So that two answers for this question. So this is one of the 1st 1 of the possible answer. The second possible answer is given Estes. So that's all for this question.

Hi friends for for frankly an elastic realism in the center of mass frame. Find a kind of technology of colliding system become zero. Hence initial kind of technology in center of mass free completely turns into yeah, internal energy right of the deformed body. Yeah. So you can write jewish culture. The initiative half mil. We want my new hospital. The square not from contribution of energy. Delta T. Is called to minus Q. Of new be even minus veto managed to re square. Mhm. In the lab frame. Same result is adopted. It's Delta T. Is going to have and when we were and to be too oh well, uh I'm unplugging them too -1. I'm a be one square half of them too. B two square which turned up as half milk. Even my next veto. Holy square, that's all. Thanks for watching it.

Okay, so here, initially bullet off mass M with speed V is colliding with em. So we have initial momentum. M v. It was final moment. Um yeah, v finals. So we find this is the initial theory initially equals m plus m over m can three final now it will go in this projectile motion. This ranges D on this height is age. So we have got h equals half G t square Because the initial rise on the spirit zero so t equals swelled off to age over G. So range de comes out to be the final times t equals. Okay, Sophie, final equals Do you have a t So we have the initial equals m plus m over em. That's the final. Which is D over tea. Yeah, Our thesis is Empress em over em. Times Square, load off. G B squared over to age.

In this question. We have given to balls and we have given its initial and final velocities and we have to tell the type of this call is um So first of all, I'm writing the given data for the execution. We are given that both have same mass, that is everyone is equal to M2 is equal so that is two kg. And we have given the velocities that is human is equal to two m/s and your two is equal to three m per second. This is the velocity before the collision. And we have given that we want is equal to two m per second and we do is equal to one m per second. This is the velocity after the collision. So we can say that if we calculate the kinetic energy before the coalition and kinetic energy after the collision. So we can say that before coalition. The kinetic energy is equal to disease half into and when you wanna square plus half into em to you to square. So this will comes out to be one by two and two. This is two and +22 square plus one by two and two. This is two and +23 square. So this is equal to 13 june similarly, we can say that kinetic energy after collision is equal to this is half and when we wanna square plus half and two Way to square. This will comes out to be one x 2 and two. This is two into 2 square plus one x 2 and two. This is two into one square and this is equals two, five jewels. As we can say that the initial and final kinetic energies are not same, so there is a loss in kinetic energy. We can say that since the kinetic energies are not the same, then the collision is any lasting collision. So this is the answer, and for that option is the correct choice. Thank you.


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