A ball with a mass of 0.620 k g is initially at rest. It is struck by a second b

Question

A ball with a mass of 0.620kg is initially at rest. It is struck by a second ball having a mass of 0.405kg , initially moving with a velocity of 0.260m/s toward the right along the x axis. After the collision, the 0.405kg ball has a velocity of 0.210m/s at an angle of 35.9%u2218 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 0.620kg ball after the collision?

What is the direction of the velocity of the 0.620kg 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

Part A)
Initial momentum is (.405)(.26) = .1053 kg m/s only in the x

After the collision, the .405 kg ball has x and y components

x component = (.405)(.210)(cos 35.9) = .06889 kg m/s

y component = (.405)(.210)(sin 35.9) = .04987 kg m/s

In the x, the momentum of the .620 kg mass is found by conserving momentum in the x

(.1053) = (.06889) + (.620)(v)

v = .05873 m/s

In the y, then, the .620 kg ball must have the same momentum in the opposite direction of the .405 kg mass

(.04987) = (.620)(v)

v = .08044 m/s

The magnitude of the velocity is found from the pythagorean theroem

v^2 = (.05873)^2 + (.08044)^2

v = .0996 m/s

Part B)

Direction is found from the tangent formula

tan (angle) = .08044/.05873

angle = 53.9 degrees below the x axis in the fourth quadrant

Part C)

Initial KE = (.5)(.405)(.26)^2 = .013689 J

Final KE = (.5)(.405)(.21)^2 + (.5)(.62)(.0996)^2= .012 J

The change is .013689 - .012 = 1.68 X 10^-3 J lost

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