Spheres A (0.020 kg) B (0.030 kg) and C (0.050 kg) are approaching the origin as

Question

Spheres A (0.020 kg) B (0.030 kg) and C (0.050 kg) are approaching the origin as they slide on a frictionless air table (see figure). 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. All three then travel at a speed of 0.50 m/s in the +x direction. (a) Find the magnitude and direction (in terms an angle CCW from +x) of the initial velocity of C. (b) Find the change in the kinetic energy of the system in the collision? Answer: (a) 1.77 m/s, +8.44 degree; (b) K= -0.092 J

Explanation / Answer

here,

vB = - 0.5 * ( cos (60) i + sin(60) j) m/s

vB = - 0.25 i m/s - 0.43 j m/s

vA = - 1.5 m/s

vC = x( cos(theta) i + sin(theta) j)

vr = 0.5 m/s

conservation of momentum along x axis

0.02 * (-1.5 ) + 0.03 *( -0.25) + 0.05 * ( x*cos(theta)) = (0.1 * 0.5) ......(1)

using conservation of momentum along y axis

     0.03 *( -0.43) + 0.05 * ( x*sin(theta)) = 0 ....(2)

     from equation (1 ) and (2)

     x = 1.77 m/s

     theta = 8.4 degree counter clockwise from x axis

     (b)

     change in kinetic energy,
     KE = ( 0.5 *ma*va^2 + 0.5 * mb * vb^2 + 0.5 * mc*vc^2) - 0.5 * ( ma + mb + mc)* vr^2

     KE =    0.5 * ( 0.1)* 0.5^2 - ( 0.5 *0.02*1.5^2 + 0.5 * 0.03 * 0.5^2 + 0.5 * 0.05*1.77^2) +

     KE = - 0.092 J

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