Colliding BallsDate: 04/28/98 at 23:10:11 From: Amy de Garavilla Subject: Conservation of kinetic energy Our physics teacher assigned us the following problem: A 5 kg ball is attached to a pendulum with a string of length 5 m. The pendulum is let go from a 45 degree angle causing the 5 kg ball to collide with an adjacent 3 kg ball. The 3 kg ball, we'll call it ball x, falls a certain distance. Upon landing it continues its path in a straight line and goes up and around a loop (as in a roller coaster.) We are asked to determine the maximum diameter of the loop. The only other information given is the distance from the edge of the area to the midpoint of the loop, which is 20 m. The coefficient of friction for this same area is 20 N. The only hint given by the teacher was "Think conservation of kinetic energy." If and how do we apply that to this problem? I am having trouble overall with this problem. I would appreciate it if you could you tell me how to go about solving for the maximum diameter of the loop so I can get started in the right direction? -Amy de Garavilla Date: 04/30/98 at 19:50:37 From: Doctor Ujjwal Subject: Re: Conservation of kinetic energy Dear Amy, This indeed looks like a very interesting problem. Unfortunately there are lots of 'missing links', but I can lay out some steps that can help you solve it (provided the missing data are filled in). There are many events occuring in this problem. Each is governed by different laws of physics. Event 1 ------- A ball raised to a certain height 'picks up speed' by converting its potential energy (PE = mgh = 5 * 9.81 * 5 * (1 - cos 45)) into kinetic energy (KE = mv^2/2). According to CONSERVATION OF ENERGY we can equate the two to get the velocity v of the 5 kg ball at the time of impact: v = square root (2gh) = square root (2 * 9.81 * 5 * (1 - cos 45)) = 5.36 m/s Event 2 ------- The two balls collide. This interaction can be analysed using CONSERVATION OF MOMENTUM and CONSERVATION OF KINETIC ENERGY (Assuming an 'elastic' collision). If: m1, m2 = masses of the balls u1, u2 = their initial velocities respectively v1, v2 = their final velocities respectively then: m1u1 + m2u2 = m1v1 + m2v2 Substitute in: m1 = 5, m2 = 3, u1 = 5.36, u2 = 0 The other equation that helps us here comes from the property of elastic collisions. In simple words it means the balls will fly apart as fast they came together! Thus the difference of their velocities before and after the collision is the same. Event 3 ------- The second ball falls through some distance. It is important to know how much distance. Because the greater this distance the higher the speed it gains. (Just like event 1 above) Event 4 ------- It travels some distance along a straight line. Again the distance is important, because you are now 'spending' the kinetic energy to 'fight against' the friction. By the way, the coefficient of friction is not related to area, is always less than 1, and has no units. So it cannot be 40 N (Newtons?). Maybe that was a typo. Event 5 ------- The second ball 'loops a loop'. Assuming the ball makes it to the top of the loop of radius R, two statements can be made: 1. the centripetal force mv^2/r must be at least equal g 2. the potential energy of the ball is 2mgR At this stage it is important to know whether the loop also has some friction. In that case the problem can be pretty involved (but still solvable). These are only some broad brush strokes. Depending upon what more is known, the depth of analysis can be determined. Do write us back with the missing data. It sure promises to be a fun problem to solve! Good luck! -Doctor Ujjwal Rane, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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