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), and C (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 and 0.50 .m/s All three spheres arrive at the origin at the same time and stick together.

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?

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

initial momentum of A = 0.020 ( - 1.50 i) =- 0.030 i kg .m/s

initial momentum of B = 0.030 x0.50 ( -cos60 i - sin60j) = -0.0075i - 0.013 j kg m/s

initial momentum of C = 0.050 v ( cos@i + sin@j )

using momentum conservatiion,

(-0.030 - 0.0075 + 0.050vcos@) i + (-0.013 + 0.05vsin@) j = (0.02 + 0.03 +0.05) x 0.50i   + 0j

-0.013 + 0.05vsin@ = 0

vsin@ = 0.26 ,,,,,,,, (i)

-0.030 - 0.0075 + 0.050vcos@ = 0.50 x 0.10

v cos@ = 1.75 ..........(ii)

(i) / (ii)

tan@ = 0.26 / 1.75

@ = 8.45 deg

v = 1.77 m/s

vx = 1.77cos8.45 = 1.75 m/s ..................Ans

[ (0.020 x 1.50^2 + 0.03 x 0.50^2 + 0.05 x 1.77^2 )/ 2 ] - [ (0.02 +0.03 +0.04) 0.50^2 /2 ]

= 0.092 J

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