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VAN PLEASE HELP ME WITH THIS QUESTION.
LLOYD ROLLS A 7.0-KG BOWLING BALL DOWN THE ALLEY FOR THE LEAGUE CHAMPIONSHIP. ONE PIN IS STILL STANDING, AND LLOYD HITS IT HEAD-ON WITH A VELOCITY OF 9.0 M/S. THE 2.0-KG PIN ACQIRES A FORWARD VELOCITY OF 14.0 M/S. WHAT IS THE NEW VELOCITY OF THE BOWLING BALL?
- HOWARD WILLIAMSON (age 17)
TRENTON, NJ , U.S.A
Well, we dont want to do your homework, but we can give you the idea to get started.
This is a conservation of momentum problem:
P(before) = P(after)
m1v1 = m1v2 + m2v3,
Where m1 is the mass of the bowling ball and m2 is the mass of the
pin. v1 is the velocity of the ball before the collision, v2 is the
velocity of the ball afterward and the v3 is the velocity of the pin
That should get you started...
(republished on 07/13/06)
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