By calculating numerical quantities for a multiparticle system, one can get a co

Question

By calculating numerical quantities for a multiparticle system, one can get a concrete sense of the meaning of the relationships P vector_ Sys = M_ tot v vector CM and K_ tot = K_ trans + K_ rel. Consider an object consisting of two balls connected by a spring, whose stiffness is 580 N/m. The object has been thrown through the air and is rotating and vibrating as it moves. At a particular instant the spring is stretched 0.35 m, and the two balls at the ends of the spring have the following masses and velocities: Here is a way to check your result for K_ rel. The velocity of a particle relative to the center of mass is calculated by subtracting v vector cm from the particle's velocity. To take a simple example, if you're riding in a car that's moving with V_ CM, X = 20 m/s, and You throw a ball with v_ rel, X = 35 m/s, relative to the car, a bystander on the ground sees the ball moving with v_ x = 55 m/s. So v vector = V_ CM + V_ rel, and therefore we have V_ rel = V - V_ CM. Calculate V_ rel = V - V_CM for each mass and calculate the corresponding K_ rel. Compare with the result you obtained in part (e).

Explanation / Answer

PART A

Psys = m1v1+m2v2 = 7*(6,13)+5*(-4,10) = (22,141)

PART B

vcm = (1/M) Psys = (1/12)*(22,41) = (1,833;11,75)

PART C

Total kinetic energy = Ktot = 0,5*(m1v12+m2v22) = 0,5(7*205 + 5*116) = 1007,5 J

v12 = 62+132 = 205 m/s

v22 = 42+102 = 116 m/s

PART D

Ktrans = 0,5*M*vcm2 = 0,5*12*15,03 = 90,18 J

vcm2 = 1,8332+11,752 = 15,03 m/s

PART E

Krel = Ktot - Ktrans = 1007,5 - 90,18 = 917,32 J

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