A ball with a mass of 0.580 kg is initially at rest. It is struck by a second ba

Question

A ball with a mass of 0.580 kg is initially at rest. It is struck by a second ball having a mass of 0.395 kg , initially moving with a velocity of 0.275 m/s toward the right along the x axis. After the collision, the 0.395 kg ball has a velocity of 0.205 m/s at an angle of 36.3 above the x axis in the first quadrant. Both balls move on a frictionless, horizontal surface.

What is the magnitude of the velocity of the 0.580 kg ball after the collision?

What is the direction of the velocity of the 0.580 kg ball after the collision?

What is the change in the total kinetic energy of the two balls as a result of the collision?

Explanation / Answer

Appying momentum conservation,

(0.580 x0) + (0.395 x 0.275i) = (0.395 x 0.205 (cos36.3i + sin36.3j)) + 0.580 v

v = 0.075i - 0.048j m/s

magnitude = sqrt(0.075^2 + 0.048^2) = 0.089 m/s ....Ans

direction = tan^-1(0.048 / 0.075) = 32.6 deg below the +ve x axis .............Ans

Ki = 0.395 x 0.275^2 /2 =0.01494 J

Kf= (0.395 x 0.205^2 / 2) + (0.580 x 0.089^2 / 2)

Kf = 0.010597 J

change in KE = Kf - Ki = - 4.34 x 10^-3 J ........Ans

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