A ball with a mass of 0.610kg is initially at rest. It is struck by a second bal

Question

A ball with a mass of 0.610kg is initially at rest. It is struck by a second ball having a mass of 0.430kg , initially moving with a velocity of 0.220 m/s toward the right along the axis. After the collision, the 0.430kg ball has a velocity of 0.215 at an angle of 37.1m/s above the axis in the first quadrant. Both balls move on a frictionless, horizontal surface.

A)What is the magnitude of the velocity of the 0.610 ball after the collision?


B)What is the direction of the velocity of the 0.610 ball after the collision? _____Below the x-axis in the forth quadrant


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

Explanation / Answer

conserving momentum in the horizontal direction....
0.43*0.22 = 0.430*0.215*cos37.1 + 0.610*vx
So, vx = 0.0342m/s
conserving momentum in the vertical direction.....
0 = 0.43*0.215*sin37.1 + 0.610*vy
vy = -0.09142m/s

so magnitude = root[vx^2 + vy^2] = 0.0976m/s
it is tan-1[0.09142/0.0342] angle below the x-axis i.e 69.489o

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