Spheres A (mass 0.020 kg), B (mass 0.030 kg), and C (mass 0.050 kg), are each ap

Question

Spheres A (mass 0.020 kg), B (mass 0.030 kg), andC (mass 0.050 kg), are each approaching the origin as they slide on a frictionless air table (Figure 1) . The initial velocities of A and B are 1.50 m/s and0.50 .m/s All three spheres arrive at the origin at the same time and stick together.

A) What must the x-component of the initial velocity of C be if all three objects are to end up moving at0.50 m/s in the +x-direction after the collision?

B) What must the y-component of the initial velocity of C be if all three objects are to end up moving at0.50 m/s in the +x-direction after the collision?

C) If C has the velocity found in parts (A) and (B), what is the change in the kinetic energy of the system of three spheres as a result of the collision?

Explanation / Answer

along x axis

initial momentum = final momentum

(ma*vaix) + (mb*vbix)+(mc*vcix) = (ma+mb+mc)*Vx

-(0.02*1.5) - (0.03*0.5*cos60) + (0.05*vcix) = (0.02+0.03+0.05)*0.5

Vcix = 1.75 m/s

along y axis

initial momentum = final momentum

(ma*vaiy) + (mb*vbiy)+(mc*vciy) = (ma+mb+mc)*Vy

-(0.02*0) - (0.03*0.5*sin60) + (0.05*vcix) = (0.02+0.03+0.05)*0

Vcix = 0.26 m/s

part C

Vc = sqrt(vcix^2 + vciy^2) = 1.77

KEi = 0.5*ma*va^2 + 0.5*mb*vb^2 + 0.5*mc*vc^2

KEi = (0.5*0.02*1.5*1.5)+(0.5*0.03*0.5*0.5) +(0.5*0.05*1.77*1.77)

= 0.1045725 J

KEf = 0.5*(0.02+0.03+0.05)*(0.5*0.5) = 0.0125 J

dK = K2 - K1 = -0.0920725 J

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