A ball with a mass of 0.605 is initially at rest. It is struck by a second ball

Question

A ball with a mass of 0.605 is initially at rest. It is struck by a second ball having a mass of 0.405 , initially moving with a velocity of 0.255 toward the right along the axis. After the collision, the 0.405 ball has a velocity of 0.205 at an angle of 37.0 above the axis in the first quadrant. Both balls move on a frictionless, horizontal surface.
Part A: What is the magnitude of the velocity of the 0.605 ball after the collision?
Part B:What is the direction of the velocity of the 0.605 ball after the collision?
Part C: What is the change in the total kinetic energy of the two balls as a result of the collision?

Explanation / Answer

a)

0.405 * 0.255 = 0.405 * 0.205 * cos37 + 0.605 * Vx

==> 0.405 * 0.255 - 0.405 * 0.205 * 0.7986355 + 0.605 * Vx

==> Vx = 0.061105

0.605 * Vy = 0.405 * 0.205 * sin37

==> Vy = 0.082588

V = (0.061105*0.061105+0.082588*0.082588)

V = 0.103

b)

= 360 - Arctan(0.082588/0.061105) =

=306 degrees    or =53.5 degrees

c)

initial = 0.5 * (0.405*0.255*0.255) = 0.013168

final = 0.5 * (0.405*0.205*0.205 + 0.605*0.10274*0.10274) = 0.011703

0.013168 - 0.011703 = 1.47e-3 J

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