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 0.405 kg , initially moving with a velocity of 0.230 m/s toward the right along the x axis. After the collision, the 0.405 kg ball has a velocity of 0.210 m/s at an angle of 37.8 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

using momentum conservation,

0.580 x 0 + 0.405 x 0.230i = 0.450 x 0.210(cos37.8i + sin37.8j) + 0.580v(cos@i - sin@j)

along j vector,

0.450 x 0.210 x sin37.8 - 0.580x vsin@ = 0

vsin@ = 0.0999 .........(i)

along i vector,

0.405 x 0.230 = 0.450 x 0.210 x cos37.8 + 0.580x vcos@

v cos@ = 0.0319 ......(ii)

(i) / (ii) =>

tan@ = 0.0999/ 0.0319

@ = 72.29 deg below

v = 0.0319/cos72.29 = 0.105 m/s   .........Ans

change in KE= final KE - initial KE

       = (0.580 x 0.105^2 /2 + 0.405 x 0.210^2 / 2) - ( 0.405 x 0.230^2 / 2 + 0 )

      = 1.41 x 10^-5 J

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