What are the physics involved when a ball rolls down a halfpipe?
470687
At the top of the halfpipe, the ball is just sitting there with potential energy stored up (PE = mgh, where m = mass, g = acceleartion due to gravity, and h = height).  As the ball rolls downward, its PE lessens and its kinetic energy increases (KE = 1/2(mv^2), where v = velocity of the ball).  At the bottom of the halfpipe, all PE is gone and KE is at maximum.  As the ball goes up the other side, the KE decreases and the PE increases.  In a perfectly frictionless halfpipe, the ball would return to the same height as it was before it started rolling, and the KE would go to 0 and the PE would be at its maximum.\n\nIf you just dropped the ball (no halfpipe - just a freefall), the KE just before impact would be the same as the PE before dropping.  The two are inversely proportional - as one goes down, the other goes up.\n\nNow if you knew the exact shape of the halfpipe, you could start plugging in numbers to find the PE and KE of the ball at various points.
1010100
ok, so if we're assuming that it was a head on elastic collision, and all the energy was transferred then there are two possible situations: if the 25lb ball was sliding then the total energy transferred will be equal to the amount it had originally (since we assumed elastic collision) that is equal to 1/2mv^2...where mass would usually be measured using the metric system. \nif the ball was rolling then it has not only translational energy (1/2mv^2) but also rotational energy which is equal to 1/2Iw^2 where I = inertia of a ball...if it is a solid sphere the moment of inertia is equal to 2/5mr^2 (m = mass, r = radius) and w (omega) = rotational speed in radians per second...which is equal to velocity/radius. so the amount of energy transferred would be equal to 1/2Iw^2 + 1/2mv^2.
591806
No, this is not perpetual motion, and the ball would not make it up the other side to come out again.\n\nThis is due to the fact that there is air resistance which would slow the ball down, even if it is very aerodynamic.  It would hit terminal velocity rather quickly, and not be able to accelerate any further.  So, in its trip back "up" the other side, it will not have nearly enough velocity to make it all the way up.\n\nThe friction with the air would eventually cause the ball to come to a stop, at a point roughly midway between the ends of the pipe (at the center of gravity).
707702
(2) You need to consider conservation of energy. The energy of the ball at point a is\n\nm*g*Ha + 0.5*m*Va^2\n\nwhich, neglecting losses, is completely converted to kinetic energy at point b. Equating them gives\n\n0.5*m*Vb^2 = m*g*Ha + 0.5*m*Va^2\n\nand solving for Vb\n\nVb = sqrt(2*g*Ha + Va^2)\n\n(3) Same considerations, however when the ball reaches its maximum height (Hm) all of the energy is potential.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hm\n\nSolving for Hm\n\nHm = Ha + 0.5*Va^2/g\n\n(4) If the block is moving at a constant speed up the ramp due to the force F then summing the forces gives\n\nF - m*g*sin(q) - Fr = 0\n\nwhere q is the angle of the ramp, m is the mass of the block, g is the gravitational acceleration, and Fr is the friction force.\n\nSolving for Fr\n\nFr = F - m*g*sin(q)\n\n(5) Again, conservation of energy, but this time at point c the ball has both kinetic and potential energy.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hc + 0.5*m*Vc^2\n\nSolcing for Vc\n\nVc = sqrt(2*g*(Ha - Hc) + Va^2)
438060
Without regard to air resistance you solve the problem as such.\n\nv(f) = final velocity\nv(i) = initial velocity\na = acceleration\nt = time\nx = displacement (distance)\n\nWhen you throw the ball up vertically, you know several things:\n1) v(i) = 30 m/s\n2) v(f) = 0 m/s\n3) a = -9.8 m/s^2 (this is because acceleration due to gravity goes in the opposite direction of the object)\n\nSolve for x with this equation:\nv(f)^2 = v(i)^2 + 2ax\n0^2 = (30 m/s)^2 + 2(-9.8 m/s^2)x\n0 = 900 m^2/s^2 - (19.6 m/s^2)x\n(19.6 m/s^2)x = 900 m^2/s^2\nx = (900 m^2/s^2)/(19.6 m/s^2)\nx = 45.9 m (rounded off)\n\nNow you know how high the ball went.  The time it takes to reach this is equal to the time it takes to get back down (assuming that it is the same distance up and down).\n\nTo solve for t, use this equation:\n\nv(f) = v(i) + at\n\n0 = 30 + (-9.8)t\n9.8t = 30\nt = 30/9.8\nt = 3.06 s  this is going up to max height.\n\nto check if it will be the same time going down, use this equation:\n\nx = v(i)t + (1/2)at^2\nv(i) = 0 (because when you reached max height, your velocity will be zero)\na = 9.8 because your ball is travelling in the direction of the acceleration.\n45.9 = 0 + (1/2)(9.8)t^2\nt^2 = 45.9/4.9 = 9.367\nt = 3.06 s\n\ndone
955845
The object has kinetic energy (1/2)*m*v^2 when first at x=0. The work friction does consumes all that energy as the spring compresses and pushes the object back to x=0. Let the distance the spring is compressed be x1. Then \n(1/2)*m*v^2 = Ff*2*x1 = mu*m*g*2*x1\n\nWhen the spring is compressed to x1, the friction has decreased the energy in the system by half and the remaining half is all stored in the spring compression because the object has come to a stop - about to start the rebound. So\nFspring*x1 = k*x1^2 = (1/4)m*v^2\nIf you solve the 1st equation for x1 and substitute the result into the 2nd, you should get the required result.
793793
the thing this problem is concerned is the vertical component of the projectile.\n\nsince the dart is thrown horizontally, the initial vertical velocity is zero.\n\nthe vertical velocity of the dart just before it hits the dart board is\n\nv^2 = u^2 + 2gs\nv^2 = 0 + 2(9.8)(0.32)\nv^2 = 6.272\nv = 2.50 m/s\n\nthe time it took the dart to hit the dart board is given by\n\nv = u + gt\n2.50 = 0 + 9.8t\nt = 0.255 sec\n\n\nnow, we have the time of travel of the dart, we can use the formula  d = vt  to find the horizontal distance\n(the horizontal velocity remained constant throughout the flight)\n\nd = vt\nd = 12.4m/s * 0.255s\nd = 3.169 m\n\n\ntherefore the dart player is standing  3.169 meters away from the board...\n\nthere ya go...
914856
OK as the ball is swung around in a horizontal circle the scale inside the tube is measuring the tension on the string. By knowing the time to complete one revolution  linear velocity can be found from v=r*omega(angular velocity). in the second part the v is given and needs to be converted into radians/second by using 3.7=.5*omega where omega=3.7/5=.74rad/s then using the mass radius and angular velocity you can find the tension in the string. With a= acceleration (into center) omega^2*r=.74^2*.5=.2738m/s^2and using f=ma the tension in the string will equal f which equals .2*.2738=.0548newtons Unless i have a silly math mistake somewhere.
13755
im going to assume this ia an elastic collision (when the objects bounce off of each other).\nyou just need the formula for conservation of momentum. p=mv (momentum = mass times velocity)  conservation of momentum says that (assuming no friction) the total momentum before the collision would equal the total momentum after the collision.  in other words mv=mv.  now, there are two balls so your formula looks like this:\n mv+mv = mv+mv.  the reason it has 2 "mv" 's  on each side, is because there are two balls.  now the second mv on the first side can be taken out since the velocity is 0, so mass times velocity would be 0. that means mv + 0 = mv + mv  so the mass times velocity of the first ball before the collision equals the mass times velocity of the first ball after plus the mass times velocity of the second ball after.  this would look like this:\nmv= mv + mv\n(1kg)(10m/s)=(1kg)v1 + (2kg)v2 (where the 1 and 2 just represent the velocities of the first and second balls respectively.)\nso basically:\n10=v1 + 2v2\nyou would need one of the velocities in order to solve for the other. there has to be more to the problem to either give you one of these velocities, or enough information to solve for it somehow... the balls would move in opposite directions.\n\nhowever, we can solve this problem if it was an inelastic collision(meaning the balls stick together).  if this were the case, then (if this is a problem for class) it probably would have said it was inelastic and it probably isnt because balls generally bounce off of each other unless they're sticky or balls of clay or something, but lets say it was inelastic just in case so i can solve it this way for you.  if it was inelastic then mv + mv=mv so the mv of the first ball plus the mv of the second ball(which is zero like before because the ball is at rest so v is zero) equals the combined masses of the balls plus the new velocity. so mv + 0 = mv, because we have only one object(because they move as one object if they stick).  the combined masses of the balls is 1kg + 2kg = 3kg so now:\nmv=mv\n(1kg)(10m/s) = (3kg)v\nsolving for v we get 10/3 or 3.3333.  they would be moving in the same direction as the initial ball.
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.
64466
Yes! There is centripetal acceleration on the ball.  It is actually accelerating in towards the center, but other forces keep it on the edge.
921378
Projectiles without the presence of air pressure creates a parabolic trajectory (path taken by a projectile).\n\n^_^
892916
I think the key here is that the objects will be rolling down the plane; thus the gravitational potential energy they have is not only converted into translational kinetic energy, but also into rotational kinetic energy.\n\nAnyway, I would approach this from energy considerations.  Start by calculating the amount of potential energy the objects will lose by falling a distance of 3 m.  (For both, this will simply be PE = m g h).  Equate this to the total amount of kinetic energy the objects have at the bottom of the ramp (translational plus rotational, or 0.5 m v^2 + 0.5 I w^2 where I is the moment of inertia, and w is the angular velocity.)\n\nSo, you need to know the moment of inertia for a disk and a hoop.  The hoop I know off the top of my head, I(hoop) = m R^2.  The disk, I looked up to find I(disk)=0.25 m R^2.\n\nFinally, recognize that as they do not slip, you can relate the angular velocity w, to the translational velocity v:  v=2 Pi R w.\nOnce you do this, you find that the kinetic energy depends only on the velocity v, so we have one equation with one unknown.  Just solve for v with the two different values of I, and I think you ought to find that the disk will be moving faster than the hoop.\n\nSorry for being long winded.... but I hope it helps.
401145
through the ball with a spin
896561
If the object starts out at a height h0 and is dropped, then all the potential energy is converted in kinetic energy\n\nm*g*h0 = 0.5*m*v0^2\n\nso that at impact the object is moving at a speed v0\n\nv0 = sqrt(2*g*h0)\n\nAfter bouncing off the surface the object is moving with a speed v1\n\nv1 = e*v0\n\nwhere e is the coefficient of restitution. In a purely elastic collision e=1, but in the real world e<1. Using the conservation of energy again\n\n0.5*m*v1^2 = m*g*h1\n\nwhere h1 is the new height to which the object bounces.\n\nh1 = 0.5*e^2*v0^2/g\n\nAt moment of next impact the object is moving at speed v1, but rebounds at speed v2\n\nv2 = e*v1 = e^2*v0\n\nThe height h2 is then\n\nh2 = 0.5*e^2*v1^2/g = 0.5*e^4*v0^2/g\n\nThis continues for however many bounces you wish. Recall now that v0^2 = 2*g*h0 so that the heights are\n\nh0 = e^0*h0 (shown only for consistency with the following)\nh1 = e^2*h0\nh2 = e^4*h0\nh3 = e^6*h0\n...\nhn = e^(2n)*h0
728073
At faster speeds you have turbulent flow, with boundary layer separation on the trailing edge. At slower speeds you have laminar flow with no boundary layer separation. Around the transition funny stuff happens to the flow where it can't quite decide what to do ( e.g. in flow around a circular cylinder, there is the Strouhal effect).  So - and I'm guessing here - that as the ball slows and the boundary layer starts to reattach, that it doesn't happen quite symmetrically, and thus imparts a torque to the ball, making it spin.\n\nThe ball is round - why wouldn't it happen symmetrically? I'd guess that some small imperfection in the air flow, possibly due to the seams in ball, gets amplified in a nonlinear manner in the air flow.
1033468
2.36 m s^-1 to 2 d p\n\nWe split the launch velocity of the ball into a vertical and a horizontal component.\nHorizontal component = 19.8 * cos(49.1) = 12.9639 m/s\nVertical component = 19.8 * sin(49.1) = 14.9659 m/s\n\nGravity accelerates the ball vertically with the acceleration -9.8 m s^-2 (negative because the pull of gravity is downwards while the ball is moving upwards to start with)\n\nThe vertical distance from the ball to the ground after time t is given by the formula s = u * t + (g * t^2) / 2 where: u = initial velocity, g = acceleration of gravity, t = time elapsed\n\nWhen the ball returns to the ground, s = 0\nTherefore the time taken for the ball to return to the ground is given by\nu * t + (g * t^2)/2 = 0 Substitute values\n14.9659 * t - (9.8 / 2) * t^2 = 0 Remove brackets, divide by t, rearrange\n4.9 * t = 14.9659\nt = 3.05427 s\n\nDuring that time the ball will travel horizontally \ns = u * t where s is distance in m, u is velocity in m s^-1, t is time in sec\nDistance travelled horizontally is 12.9639 * 3.05427 = 39.5953 m\n\nThe player has to run 46.8 - 39.5953 m = 7.20475 m in 3.05427 s\nHis speed must therefore be 2.35891 m s^-1