the head of a golf club (200 g) is traveling at 55 m/s just before it strikes a 46 g ball at rest on a tee. After collision, the club head travels in the same direction at 30 m/s.
What is the speed of the ball just after impact?? explain??
2007-10-16 07:31:38 · 2 answers · asked by Captain Whiskerboy Litterbox 3 in Science & Mathematics ➔ Physics
I think the idea here is that the golf club has lost momentum because of the impact:
club's momentum just before impact = (55m/s)(200g)
club's momentum just after impact = (30m/s)(200g)
Because momentum is conserved, the club's lost momentum must have been transferred to the golf ball.
momentum lost by club =
momentum transferred to golf ball = (46g)(v)
From that, you should be able to solve for "v".
2007-10-16 07:39:44 · answer #1 · answered by RickB 7 · 0⤊ 0⤋
This is a peculiar problem. What you'll find is that the ball is going 108.7 m/s. While this satisfies conservation of momentum it violates conservation of energy. In a perfectly elastic collision, the relative speeds of the two bodies remains the same through the impact, and their post-impact speeds are distributed such that both momentum and energy are conserved. In this case the relative speed increases from 55 to 79 m/s which implies a coefficient of restitution (CR) of about 1.43. No real impact has a CR > 1. Enter the numbers in the calculator on the ref. page and you'll see.
Note: Yahoo currently has a bug that prevents posting the URL unless a space is inserted. (You get a "999" error.) To reach the ref., paste it into your browser's address bar and remove the space before "edu".
2007-10-16 14:59:45 · answer #2 · answered by kirchwey 7 · 0⤊ 0⤋