hw conceptual problem Consider a perfectly inelastic collision between a car and a large truck.?
960455
1.  Aside from the fact that the superhero is, in the first place,  defying the laws of gravity there is one other major flaw in this scenario.  For every action there is an opposite and equal reaction.  The superhero has imparted momentum to the piano and yet we do not see the "opposite and equal reaction".   Perhaps you have seen the true, real life, effect in other movies you may have seen.  Movies in which cannon are being fired.  When the cannon fires it does what?  It recoils.  Armies and Navies for that matter take great pains to control the flight of the cannon upon recoil.  Our superhero seems to be immune to recoil, immune to the conservation of momentum and thereby immune to the laws of physics.  I call "Foul"!\n\n2.  If the truck is at rest when struck by the car, then the car loses KE while the truck gains KE.  If the truck is in motion when struck by the car, in an inelastic collision, meaning the car an truck become entangled and inseparable, then the truck most certainly loses the most KE.  KE= 1/2 * m* v^2.  Velocity (v) is the key here, velocity and it's relationship to momentum (p=m * v; p - momentum). \n\nIn an inelastic collision, conservation of energy is maintained but conservation of KE is not.  (Some of the kenetic energy will be converted to the energy of deformation, sound of the collision)  The Truck will retain most of it's velocity whlie the lower mass car will accelerate largely in what ever direction the truck was headed, most likely causing a net loss of absolute velocity and therefore a loss of kenetic energy for the car.  But the mass of the truck is much larger than the mass of the car.  So the answer to the original question actualy depends on the speed of the truck compared to the speed of the car before the collision.  Why?  Because the collision is inelastic.\n\n"Perfectly elastic collisions are those in which no kinetic energy is lost in the collision. Macroscopic collisions are generally inelastic and do not conserve kinetic energy, though of course the total energy is conserved. The extreme inelastic collision is one in which the colliding objects stick together after the collision, and this case may be analyzed in general terms"\n\n3.  Force and Pressure.   Pressure is simply force exerted over a specified area.  In a car crash, pressure is what kills us and in the case of the airbag, Counter-Force over a wider area is what saves us.  Pounds per square inch, Phyiscs calls that Pressure.  Pressue = Force / Area while Force = M * a^2.  We know that an object in motion tends to stay in motion.  And we know that the law of conservation of momentum guarentees the universe that what is in motion tends to stay in motion and what is at rest tends to stay at rest.  That being said, Force exerted over a given area is called Pressure.  During a collision, a human body has, prior to the collision, a velocity, a mass and therefore a momentum.  And that momentum drives that human being to "stay in motion" even after a collision occurs.  In order to decelerate the body, a force must be imparted.  Enter the airbag.  The airbag inflates.  As the body presses against the airbag, the airbag compresses.  As the air bag compresses and it's overall volume is reduced, the area over which the air pressure acts is less therefore the force / area is more causing a "push back" or Counter-Force affect on the body pressing into it.   Opposite and Equal.  As long as the airbag is strong enough to withstand the surge of pressure (force/area (pounds per sqaure inch)) then the human body will be protected by the absorption of energy that is required to deform the plastic membrane that is the "airbag".  The human frame will be allowed to more slowly decelerate from the collision.  That deceleration will be slow enough to allow the cells and musculature of the human body to maintain its functionality.
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It does not depend on the degree of the slop.\n\nThis is yet another simple conservation of energy problems that Drew doesnt want to do on his own.\nhttp://answers.yahoo.com/question/index;_ylt=Aml6Tp9rfj_6pwH6nEZzn8nsy6IX?qid=1006022212835\nOften the only think that changes in the questions (if anything at all) are the numbers.  The process used to solve it stays the same.\n\nThe intial Kinetic energy of the car is transformed into gravitational potential energy.\n\nchange in  KE = change in PE\n\nKE = 1/2 mv^2\nPE = mgh\n\nSince mass (m) is the same in both equations, it cancels out.\nSo the change in height (h) is equal to v^2 / 2g\nwhere g is the gravitation acceleration (9.81 m/s^2).
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a=F/m, lim x->inf x/x = 1. The object will accelerate.
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For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
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a=(delta V) /t = (26/24)=1.08 m/s2\n\nF= mxa,   \n\nwe know weight of car is 11420 N, we need mass\n\nm=11420 / g= 11420 / 9.81 = 1164 kg\n\nso F= 1164 x 1.08 = 1257.25 N to stop the car\n\nfriction = F/Fn =  1257.25/11420 = .11
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You can solve this problem by resolving the forces along the ramp for part a) and normal to the ramp for part b).\n\na) The net force along the ramp must be zero for the crate to move at constant velocity (ie no acceleration), so we have:\n\nFcos30 = mgsin30\nF=mgtan30\n\nb) the reaction of the ramp on the block will be normal to the ramp and equal to the sum of the imposed forces resolved normal to the ramp, so:\n\nR = Fsin30 + mgcos30
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double_nubbins has the right approach.  I will add a couple of hints: how long will it take the car to fall 19ft (remember the acceleration equation s=.5*g*t^2).  Assuming there is no air drag, the horizontal velocity of the car must be enough to cover the 52m width in that time.\n\nSpeed of the car? The horizontal speed? The falling speed?  The speed in the direction of motion?\n\nHorizontal speed: as calculated above\nFalling speed = g*t, where t is the time calculated above.\nSpeed  in the direction of motion = sqrt(horiz speed ^2 = falling speed ^2)
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If the objects are perfectly elastic (that is they don't deform), then you can use the laws of conservation of momentum and conservation of energy to solve the problem. If they (or one of them is) are plastic...that is they deform during the collision, then conservation of momentum still applies, but it is more difficult to use conservation of energy unless you know how energy is lost in the deformation process.\n\nI don't mean that to sound complicated, but that's it in a nutshell.
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Alright... physics is fun if you get the right problems :) .  \n\nThe collision is elastic... your book should have a conservation of momentum equation.  It states that p1+p2 = Ptotal\n\nYou have to use component vectors for the velocity after the collision.  A good thing to do would be to draw the problem.  It would make it a lot easier.  So use conservation of momentum.\n\nAnd there are two equations that would really help you:\n\nFinal velocity of block A= ((MB-MA)/(MA+MB))*VAinitial\n\nFinal velocity of block B= ((2MA)/(MA+MB))*VAinitial\n\nHelp with physics:\n\nALWAYS draw pictures epecially on your exams.  Get on the good side of the professor, go to a tutoring center if you have one, do many problems and memorize how to do the basic ones if you want an A on the test.
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This is an interesting question.  I think tyba has the right idea.  I would say that higher temperatures lead to a more flexible ball, which in turn will lead to a more elastic collision with the ground.\n\nJust in case you don't know:\nA perfectly elastic collision is one in which the object has exactly equal and opposite momentum after the collision (no loss of kinetic energy).  A perfectly inelastic collision is one in which all kinetic energy is converted to another type of energy.  No real collision is either one of these, but rather somewhere on the spectrum between them.\n\nIn the case of the frozen ball, I would say that most of the energy is converted to thermal energy (heat).
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No energy is lost as a result of heat or material deformation in an elastic collision.  It simplifies the mathematics by not having to keep track of non-motion energy, but it does not reflect practical results.\n\nIf automobile collisions were totally elastic, body shops would be out of business but funeral homes and hospitals would have a boom.
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Gravitational potential energy at top is V = mg(z2 - z1)\nV = 1300*9.81*(220) = 2,805,660 J\n\nKinetic energy at top is T = 1/2*m*v^2\nT = 0.5*1300*29^2 = 546,650 J\n\nTotal energy of car at top = T + V = 3,352,310 J\n\nWork done on car to get it to top is 9,000,000 J\n\nThus energy lost to non-conservative forces is:\n9,000,000 - 3,352,310 = 5,647,690 J
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When an object (like the truck) hits a solid wall then the wall forces it to stop suddenly. \n  The wall and the truck have a certain elasticity, though small. When the truck touches the wall it starts pressing on it and the wall bends very little. It bend in a very short time too. During this small period of time the object's momentum changes from "mv" to "0". The force is the ratio F= (m*v) / t. Because t is very small the force exerted is very big and it crushes the truck's front end.
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USE PRINCIPLE OF CONSERVATION OF linear MOMENTUM.\n\nearth gas lower KE because it has huge mass and huge inertia
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50 Km is the answer\n\nIf the car moved 50 km north and then 50 south its displacement would be 0\n\nif it go in a direction like sw or ne or nw then you have to use a triangle to solve its displacement
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It's tranferred to heat/friction from the impact
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I normally refrain from answering questions that have already been answered by so many people, but in this case, so many of the answers are incorrect, and it might be useful to have a single answer that clarifies what's correct and what isn't.\n\nAn object in freefall in air is subject to two forces. Gravity, which directly proportional to the mass of the object, pulls the object down. Drag acts opposite to the object's movement, and is related to the objects size, its shape, the square of its velocity, and the density of the fluid through which it's traveling. In this case, the drag force acts upwards\n\n^  drag\n|\nO  object\n|\nv  gravity\n\nThe force due to gravity is essentially constant (it increases slightly as the object gets closer to the earth, but this increase is negligible). The force due to drag is related to the size ans shape, which is constant for rigid bodies, and to the speed, which is increasing initally.\n\nThe objects will accelerate until the forces are equal. This occurs at terminal velocity. I would estimate this to be approximately 100 mph. It would take more than 300 ft to reach this speed.\n\nIn a vacuum, there is no drag force, so the acceleration due to gravity (about 9.8m/s^2 near the earth's surface) would continue to accelerate the objects until another force acted on them (e.g. hitting the ground).\n\nSimilarly sized, shaped, and weighted objects such as an egg and an orange could be expected to fall at almost the same rate.\n\nOne final note has to do with a term I used earlier. Neither an orange, nor and egg is really a "rigid object". The egg especially might experience some degree of sloshing inside which could affect its motion. This is why tanker trucks have compartments. I'm not sure if you'd notice this effect, but if you hard boil one egg, you could perform an experiment yourself.\n\nThe bottom line is that you really shouldn't notice any difference in the way the orange or the egg falls.