Spheres A (mass 0.021 kg), B (mass 0.028 kg), and C (mass 0.052 kg), are each ap

Question

Spheres A (mass 0.021 kg), B (mass 0.028 kg), and C (mass 0.052 kg), are each approaching the origin as they slide on a frictionless air table (Fig. 8-37). The initial velocities of A and B are given in the figure. All three spheres arrive at the origin at the same time and stick together.

(a) What must be the x- and y-components of the initial velocity of C if all three objects are to end up moving at 0.50 m/s in the +x-direction after the collision?
m/s (x component)
m/s (y component)
(b) If C has the velocity found in part (a), what is the change in the kinetic energy of the system of three spheres as a result of the collision?
J

Explanation / Answer

apply the law of conservstion of momentum along x axis

initial momentum = final momentum

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

-(0.021*1.5) - (0.028*0.5*cos60) + (0.052*vcix) = (0.021+0.028 +0.052)*0.5

Vcix = 1.71 m/s

similarly now along y axis s

initial momentum = final momentum

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

-(0.021*0) - (0.028*0.5*sin60) + (0.052*vciy) = (0.021 +0.028 +0.052)*0

Vciy = 0.233 m/s (ANSWER)

-------------------------------------------------

part C

Vcnet^2 = Vcix^2 + Vciy^2

Vcnst^2 = 1.71^2 + 0.233^2

Vcnet = 1.725 m/s

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

KEi = (0.5*0.021*1.5*1.5)+(0.5*0.028*0.5*0.5) + (0.5*0.052*1.725*1.725)

KEI = 0.10499 J

similarly final KE = KEf = 0.5*(0.021+0.028+0.051)*(0.5*0.5) = 0.0125 J

change in KE = dK = K2 - K1 =0.0125 - 0.10499 = -0.0925 J (ANSWER)

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