Colliding Marble Question?
817639
2 marbles.  It will be the one that was sitting alone, as well as the last of the 5.  In these perfectly elastic collisions, which we idealize for this thought experiment, the momentum is transferred without loss down the line, ending up only in the last recipient.  Similarly, if you shot 2 marbles at a single marble sitting alone, only the second of the 2 marbles would stop.  The first would keep moving, now with the one that had been sitting alone, so the quantity of moving marbles would be 2, the same as what was moving in the beginning, even though it is different individual marbles.
805410
Let me ask you a question. \n\n If I shot a bullet straight up, and then dropped a feather..  would the falling object move faster than the one going upwards?
501193
In physics a collision is said to take place \n * When two bodies physically collide against each other or\n * When the path of one body is changed by the influence of the other body.\n \nEXAMPLES :\n  We have several examples for collision\n * Marbles or carom coins colliding with each other\n * Molecules of a gas colliding with one another etc.\n\nThere are two types of collisions.\n1. ELASTIC COLLISION\n2. IN-ELASTIC COLLISION\n\n1. ELASTIC COLLISION:\n   If there is no loss of kinetic energy during a collision, it is said to be elastic collision.\nExample: collision between atomic and nuclear particles are elastic.\n\n2. INELASTIC COLLISION:\n   If there is loss of kinetic energy during a collision it is said to be in-elastic collision.\nExample: If a ball is dropped from a certain height and the ball is unable to rise completely to its original height, then it would mean that the ball has lost some kinetic energy due to collision.\n\nHEAD-ON ELASTIC COLLISION:\n  It is that elastic collision in which the colliding bodies move along the same straight line path before and after collision.
763881
Take the x-axis as the line of action of the moving ball. and suppose the second ball moved at an angle theta wrt to the line of action with certain velocity that u need to find out. Now resolve the velocity components of both balls along x-axis and y-axis before and after collision. Now conserve the momentum along both x-axis and y-axis before and after collision. solve the equations and u will know all the unknowns.\n\nFor the second question find 'e' i.e. co-effecient of restitution and depending on the value it will be elastic or inelastic.
891480
Calculate the k value for the spring.  The stone exerts a force of 8*9.8 = 78.4 N and a compression of 9 cm is\n78.4/.09 = 871 N/m. (Keep it in the MKS system ☺)\n\nWhen the spring is pushed down another 30 cm its total deflection is 39 cm and it's total potential energy is \nkx²/2 = 871(.39)²/2 = 132.5 J\n\nThe stone acquires 132.5 J of gravitational potential energy when released.\n\nSince gravitational potential is mgh, then\n8*9.8*h = 132.5 or h = 1.69meters\n\n\nDoug
112172
Only a limited answer:  For the special case where M = m, there are just 2 collisions.  After the first collision, V = 0 and v = V's initial value.  In the post-reflection collision the opposite exchange occurs.
1095641
Think of the Earth as a big blue marble. Got it? Ok. Now, think of a layer of bubble wrap around the marble. That would be the air we breathe, or the troposhpere. It is air, so it doesn't weigh much, but it does weigh something. Now think of another layer of bubble wrap on top of the first. That would be the stratosphere. Then another layer, which is the mesosphere, and then another layer, the thermosphere. Now you've got four layers of bubble wrap, on top of each other. Each one is PRESSing down on the ones beneath it, causing PRESSURE. Hope that helps.
546536
Take two objects of the same mass moving towards each other (in opposite directions) with the same speed. Let's say one object has momentum of p and the other -p. The total momentum of the system (p-p) is 0. When they collide, they both stop moving and the total momentum of the system is still 0.
879264
'Large' and 'small' probably refer to the masses of the balls, rather than the sizes. Assume one ball (with mass m1) is initially directly above the other (mass m2), and both are dropped at the same time from rest (with a slight vertical space between the two).\n\nThe lower ball hits first, bounces up and collides with the upper ball near the ground. The two  have about the same speed v = sqrt(2gH) going into this collision because both balls have fallen from approximately the same height. If the two masses were equal, then both would simply reverse direction, the top ball going upward at speed v and the bottom ball going down (for the second time) and immediately up at the same speed v --> both balls would continue up at the same speeds to the original height H.\n\nNow suppose the masses are slightly different with m2 > m1. Then the initial and final velocities (taking the upward direction as positive) for the collision are respectively:\n\nv1 = -v\nv2 = v\n\nand:\n\nv1 = v + x1\nv2 = -v + x2 \n\nwhere x1,x2 << v. Energy and momentum conservation require that:\n\n(m2-m1)v = m1(v + x1) + m2(-v + x2)\n(m1+m2)v^2 = m1(v + x1)^2 + m2(-v + x2)^2 \n= m1(v^2 + 2vx1) + m2(v^2 - 2vx2) (Approximately)\n-->\n2(m2-m1)v = m1x1 + m2x2\n0 = 2m1vx1 - 2m2vx2\n-->\nx1 = (m2/m1 - 1)v\nx2 = (1 - m1/m2)v\n\nSo the top (lighter) ball ends up with a higher speed (m2/m1)v and the lower (heavier) ball ends up with the lower speed (m1/m2)v. This is reflected by the one bouncing higher than H and the other lower than H.\n \nThere can't be more than one collision since the top ball comes out of the first (and only) collision with a higher speed than the bottom ball so the latter can never catch up to it. If the mass ratio were reversed (heavier ball on top) the situation would be much more complicated since there could be multiple collisions.
796966
Marble 1 will stop dead, marble 2 and marble 3 will move off at 7.7.5 m/s 45 degree to the right and to the left.  If you could see their trails painted up in light, it would make a capital letter Y but with that upper portion being at a right angle.
288682
The most common way to think of that would be, by lifting the marble higher.  By lifting it you give it potentiol energy that can be released by dropping it.  You could also push it into a spring.  As the spring gets tighter you are storing Potential enrgy that can be released by letting go of the marble.  A sling shot would do the same thing.
93828
I think the hardest question ever asked was asked by a scientist.  This was the question, and it literally drove him insane thinking about it:  Two large metallic orbs are traveling towards each other on a collision course.  Both are indestructable, and nothing can divert them from their path in any way.  They can never stop moving on their path for any reason.  What would happen when they collide?  Neither can be moved from the path that they are traveling, they cannot be destroyed, and they can never stop moving.
335694
For the first case A marble on a (perfectly) flat surface, will remain in that state until energy of some kind is added.\n\nFor the second case to be as close to perfect conditions you would need to have a moving object out in deep space away from other bodies, because gravity will always be an effect, and in space to avoid the friction of the air.\n\nBut in simple terms the same marble on the flat surface give a push will move until external forces stop it, friction from the air and from contact with the surface.
761342
We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.
357332
Is this trick question?
716426
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.
170531
1) V2^2=V1^2+2d*g ----> 0=47^2+2d*-9.8 ----> d=112.7 m\n2)remeber when mesuring the time we take the velocity on the Y axes and when mesuring range or distance we take velocity on the X axes got it?\nd=V1+0.5*g*t^2 and since the rock is thrawn horizontaly the velocity on the Y axes once it's thrawn is zero, by the way the decplacement would be minus cuz the rock is projecting downwards \n-3.5 m = 0 + 0.5 * -9.8 * t^2 -----> t=0.845 seconds\nd=V*t=47*0.845= 40 m\n3)this is a tough one you'd better draw it :\na.V2^2=V1^2+2d*g ----> 0= (47sin37)^2 + 2d*-9.8 ----> d=40.8m\nb.V2=V1+g*t ---> 0=47sin37+-9.8*t ---> t=2.886 seconds\nrange=V*cos37*t=47cos37=108 m\nc.V2=V1+g*t=47sin37+(-9.8*2)=8... m/sec (velocity on the Y axes)\nvelocity on the X axes is constant and equals=47cos37=37.5m/sec\nthe net velocity^2=37.5^2+8.685^2 ---->velocity of rock after 2 seconds of launch=38.5m/sec