Two discs are free to move without friction on a horizontal table. The 0.400-kg

Question

Two discs are free to move without friction on a horizontal table. The 0.400-kg disc is initially at the position (x=0, y=1.0) m, moving with velocity (v_x =4.00, v_y = 0) m/s. The 0.600-kg disc is initially at (x=1.5, y=0) m, moving with velocity (v_x =0, v_Y =6.00) m/s. The figure above displays the initial conditions for the two discs in the x, y-coordinates. The initial velocity of the center of mass of the two-disc system is {V_x, V_y) = (4.00, 6.00)m/s (V_x, V_y) = 6.00, 4.00)m/s (V_x, V_y) = (1.20, 1.20)m/s The two discs collide at some point in the x-y plane. What is the velocity of the center of mass after the collision? Less than before the collision. The same as before the collision. It depends on whether the collision is elastic or inelastic. Impossible to determine If the collision is completely inelastic, what is the magnitude of the final momentum in the system? What is the ratio of the system's kinetic energy after to kinetic energy before the collision? (K_B/K_b)

Explanation / Answer

velocity of COM

VCM = (m1V1 +m2V2)/(m1+M2)

m1 = 0.4kg V1x = 4.0 m/s V1y =0

m2= 0.6 kg   V2x = 0 V2y = 6.0 m/s

m1 and m2 have velocities along x and y hence

VCMx = (0.4*4.0+0)/(0.4+0.6) = 1.6 /s

VCMy = (0+0.6*6.0)/(0.4+0.6) = 3.6 m/s

ans: (d)

after colision it depends on the type of collision, elastic or inelastic.

momentum is conserved with inelastic collision.

momentum before collision

=   0.4*4.0 = 1.6 kg-m/s in x-direction

    0.6*6.0 = 3.6 kg-m/s in y-direction

momentum after collision magnitude = sqrt(1.62 +3.62) = 3.94 kg-m/s

KE of the system before collision = 0.5*0.4*42 +0.5*0.6*62 = 14 J

velocity of COM (magnitude)= sqrt(1.62 +3.62) = 3.94 m/s

KE after collision = 0.5*1*3.942 = 7.76 J

ratio = 7.76/14 = 0.55

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