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Launching with Ramp ( IOp] Two identical ice cubes slide down Irictionless on Ewo rmps and both are launched with the same inicial spccd of 2,00 ms into projectile ...

Question

Launching with Ramp ( IOp] Two identical ice cubes slide down Irictionless on Ewo rmps and both are launched with the same inicial spccd of 2,00 ms into projectile motion The ice cube from the blue rmp aunched horizontally and the ice cube Irom the yellow Nmp launched at an anglc of 179 bclow thc horizonaal The point cncre the ice cubes start their projectile mouon 80 cm abovc tc floor:yellow2 msblue80 cmAt which horizontal distance; from the edze of the tblelramp: does the ice cube from the yel

Launching with Ramp ( IOp] Two identical ice cubes slide down Irictionless on Ewo rmps and both are launched with the same inicial spccd of 2,00 ms into projectile motion The ice cube from the blue rmp aunched horizontally and the ice cube Irom the yellow Nmp launched at an anglc of 179 bclow thc horizonaal The point cncre the ice cubes start their projectile mouon 80 cm abovc tc floor: yellow 2 ms blue 80 cm At which horizontal distance; from the edze of the tblelramp: does the ice cube from the yellow ramp land? (8p) The ice cube from the blue ramp (select all statements that are correct); lands at thc same time a5 thc icc cubc from thc yellow rmp 40 Sp) lands thc samc distance away from the cdge of the table as the ice cubc from the ycllow rmp 40 Sp) lands on the ground with the same specd the ice cube from the yellow Nmp (p)



Answers

A projectile is launched from $\mathrm{O}$ at the foot of an inclined plane inclined at $30^{\circ}$ to the horizontal, with an initial velocity $\mathrm{u}_{1}$, at angle of projection $60^{\circ}$ to the horizontal to reach a point $B$ on the top of the incline. From the position $B$ it is launched with velocity $\mathrm{u}_{2}$ at $60^{\circ}$ to the horizontal to reach back $\mathrm{O}$ at the foot. If $\mathrm{t}_{1}$, and $\mathrm{t}_{2}$ are the times of flight in 1 st and 2 nd cases respectively, then match the following: Column I (a) $\frac{\mathrm{u}_{1}}{\mathrm{u}_{2}}$ (b) $\frac{\mathrm{t}_{1}}{\mathrm{t}_{2}}$ (c) Modulus value of acceleration normal to the plane (in $\mathrm{ms}^{-2}$ ) (d) Modulus value of acceleration parallel to the plane (in $\mathrm{ms}^{-2}$ ) Column II (p) $5 \sqrt{3}$ (q) $\frac{1}{\sqrt{2}}$ (r) $\sqrt{2}$ (s) same for both

In this problem we have given in a game of physics and skill. A reserved block with mask m sits at rest at the age of frictionless air table. Suppose this is a table like this. This is block A And here another blog Initially treaties VB The mass of blog is equal to mass of block A and Collison is elastic. Coalition is elastic so we know that in elastic collision if kind of take energy is equal, we know that you know elastic realism. Kinetic energy is equal before and after that call is, um so we know that here freak stabilised friction less. That means net external force will be equal to zero. So after the coalition yes, If blocks are identical and Collison is elastic, they interchange their velocities Like initial supposed initial blog has initial velocity V V But after collision blog will be addressed and block a has velocity will be here after the call is um say no. This vlog after collision will be go will go like this and this distance X is given to us. This X is 2 m and this height is also given Mhm. This height is equal to 1.20 m. So we know that X is equal to this is spilled. We be into time now we can find a time in projectile motion. Time is equal to two h by G. So here to into edge 1.20 divided by 9.81 So time is equal to a result. This is equal to 0.49 46 seconds. We will put time here. So we were called to X Y. Maybe this is a call to two divided by 0.4946 This is the speed off. Yeah, uh, block a after collision. This is 4.4 m per second and the same speed will be off the lobby before the call is, um so this is a better answer.

You can say the layer of ice first would be equaling to the density of 917 kilograms per cubic meter multiplied by 400 meters by 500 meters by 0.4 rather point 00 for zero meters. And this is giving us 7.34 times 10 to the fifth kilograms. This this added to the mass of the 100 stones and we have the mass of the stones equaling 100 stones at 20 kilograms each. And so we can then say that the total mass M simply would be 7.36 times 10 to the fifth kilograms. Now we're going to four part a set f equaling Dee dee would be, of course, the drag force and we're using Equation 6 14 and this would allow us to find the wind speed V along the ground. And we can say that then the wind speed V would be equaling the square root of the coefficient of kinetic friction mg. This would be divided by four multiplied by the drag coefficient for ice times. The density times the area for the ice and we can solve. Let's get a new workbook the velocities than Equalling the square root. Uh huh! 0.10 multiplied by 7.36 times 10 to the fifth kilograms multiplied by 9.8 meters per second squared. This will be divided by four multiplied by 40.2 multiplied by 1.21 kilograms per cubic meter multiplied by 400 by 500 meter squared. And this is giving us approximately 19 meters per second. We can convert 19 meters per second, multiplied by one kilometer for every 1000 meters, multiplied by 3600 seconds for every one hour. And this is giving us approximately 69 kilometers per hour. So either answer would be correct. For part a doubling for part B. We have doubling our results from Cartier. We find that the reported speed the is equaling 139 kilometers per hour and for part C, is it reasonable when we can say yes? This is reasonable. Four windstorms Ah, where a Category five hurricane we can say cat five hurricane. Those have maximum brother. Those have speeds around typically 260 meters per second Category five being, of course, the strongest hurricane. That is the end of the solution. Thank you for watching

The first part. Why component of I had what is equal to by a component of you know what? Because too bad actors and at the moment, but a nuclear direction, Uh, have no effect. Yeah, on each of them. Oh. Oh, yeah, yeah. Therefore, mhm Mm. Buy Xs behavior. Mm. Yeah. After everybody. Uh oh. Will depend only all It's by component velocity since mhm red and blue balled. Reached to the same height. Mhm. So do we. Not by a red one. It's called to be not by off. Well apart. Second part. Okay, X component B not X of redwood. It's called Do we not x off? And no one Because both covered same distance? Yeah, in horizontal direction. Oh, in the same time. Yeah, They're so thanks for

So we can say relative to the sled, the launch velocity be equal to the initial launch velocity in the X direction I had plus the Y component of the initial launch velocity Jr we know that the sleds motion is a negative direction with speed. V sub S. V sub S is a positive number. We are treating it as a speed, not a um not a velocity. So we could say that the sled velocity is actually negative Visa. S. I. So the launch velocity then this would be relative to slut here relative to the ground. We have then the X. Component of the launch velocity relative to the slud minus the speed I had plus the Y component J hat your relative to ground. So given this, we can say that the horizontal and vertical displacement again relative to the ground. And so we have X sub land minus x sub launch or the horizontal displacement Equalling the uh huh horizontal displacement B. G. Again this is all relative to the ground. This would be equal to the X. Component minus the speed multiplied by the time of flight. Why so bland minus wise of launch. This is equaling zero. You know that this would be equal to the Y component of the initial velocity multiplied by the time of flight And plus one half. Then times. Then the negative acceleration due to gravity multiplied by the time of flight quantity squared. So we can combine these two equations and we see that the horizontal displacement is going to be equal to to be an X initial. The Y initial divided by G minus to the y initial divided by G. Multiplied by the speed. The first term essentially here is a quote on is like a Y intercept essentially. And so we can say that then delta X sub B G would be equaling 2 40 minus four V sub s. This essentially implies that the initial component initial Y component of the velocity 4.07 seconds, Multiplied by 9.8 m/s squared. This would then be divided by two Equals 19.6 m/s. And we can then say that for part A The x component would be equaling two 40 meters, multiplied by 9.8 m per second squared, divided by then two Times 19.6 m/s. of course this is equally then 10 meters per second. So sure for part B we have the Y component already solved. And for part C we can say relative to the sled, we can say that then the displacement relative to the sled does not depend on the sleds speed. So here this is simply equal to the X component multiplied by the time of flight and solved this would equal 40 m, so and again for part D, this is the same exact thing as part C relative to the sled. The displacement in the horizontal direction does not depend on the sled speed. This again equals the initial, the X component of the initial velocity multiplied by the time of flight. No surprise, of course, 40 meters. That is the end of the solution. Thank you for watching.


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