A 0.20-kg billiard ball traveling at a speed of 15 m/s strikes the side rail of a pool table at an angle of 60o. If the ball rebounds at the same speed and angle, what is the change in its momentum?
2007-07-22 08:46:14 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
To split some hairs with the definition, I would say no change in the absolute momentum since there is no change in stated SPEED nor mass. There has been a perfectly elastic compression and rebound with no energy loss. The vectors could enter into a subsequent consideration.
2007-07-22 10:10:35 · answer #1 · answered by Bomba 7 · 1⤊ 0⤋
Momentum is a vector quantity (direction matters). When approaching the side of the table, its component of momentum perpendicular to the side is
+0.2x15xsin(60) (toward the side is defined as +)
After hitting the side of the table, its component of momentum perpendicular to the side is
-0.2x15xsin(60) away from the side (this direction is -)
So change (final - initial) is
-0.2x15xsin(60) - (+0.2x15xsin(60) ) = -2x0.2x15xsin(60)
= -3xsqrt(3).
Momentum parallel to the side is unchanged.
Before, momentum is 0.2x15xcos(60) in a given direction
After, momentum is 0.2x15xcos(60) in the same direction
So change is
0.2x15xcos(60) - 0.2x15xcos(60) = 0.
Therefore total change in momentum is -3xsqrt(3) where - sign indicates "away from the side".
2007-07-22 16:45:20 · answer #2 · answered by HC 1 · 0⤊ 0⤋
2*(.2*15)cos60o
2007-07-22 16:35:56 · answer #3 · answered by Heart Break Kid 2 · 0⤊ 0⤋