Newton's Cradle W/ Motion Study and Physics Analysis
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Because this is not an elastic collision, kinetic energy is lost. We can model the kinetic energy of the 5-ball system by examining the collision immediately before and after they collide: KE (before): 1/2mv^2 KE (after): 1/2(5m)(v')^2where m represents the mass of one ball, v represents the maximum velocity that the first ball has before the collision, and v' represents the maximum velocity that the 5-ball system experiences.Kinetic energy is not conserved, but according to the law of conservation of energy, total energy is conserved. You can model total energy by taking into account the potential energy due to gravity and thermal energy, where thermal energy is a general description for the frictional forces present in the system:TE (before first ball is dropped): mghTE (immediately after balls collide): 1/2(5m)(v')^2TE (when 5-ball system reaches peak): (5m)gh' +5(F_s)Lwhere h represents the difference in height between the first ball's center of mass and the rest of the balls' center of masses before the first ball is dropped, h' represents the difference in height between all of the balls' center of masses when they have instantaneously stopped and their initial position immediately after they collided with each other, F_s represents the force of static friction between a pair of balls, and L represents the distance at which the force of static friction works.We're not done though yet. We can also calculate the frequency and period of the oscillations of the 5-ball system after the collision. We do this by drawing a free body diagram of the system and creating an equation to represent the forces at play. In the tangential direction (perpendicular to the tension force), the only force acting on the 5-ball system is horizontal component of gravity:-(5m)g(sin(θ(t))) = (5m)awhere θ(t) represents the angle as a function of time between the tension gravity force.From this we can cancel the 5m on both sides and create a differential equation:d^2x/dt^2 +gsin(θ(t))=0If we assume the oscillations are small, then θ(t) is small and by using the power series representation of the sine function, we can say sin(θ(t)) ≈ θ(t). Therefore, our new equation is:d^2x/dt^2 +g(θ(t))=0Due to the sector length formula, we can say:x(t) = R(θ(t))where R is length of the thread and x(t) represents the distance that the 5-ball system travels in terms of time. Now our differential equation is in the form:d^2x/dt^2 +(g/R)x(t)=0Finally, because our differential equation is in this form, the solution is:x(t) = Acos(ωt)where A represents the maximum height of the oscillation (can also be written as h') and ω = sqrt(g/R).From this:Frequency: (1/2π)sqrt(g/R)Period: (2π) sqrt(R/g)NOTE: These calculations were all under the assumption that the mass of threads holding up the balls are negligible. Also, the force of air resistance is insignificant, so the oscillations are not damped.
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