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Transfer of kinetic energy

  • Thread starter Ry122
  • Start date
  • #1
565
2
High-speed stroboscopic photographs show that the head of a golf club of mass 210 g is traveling at 52.7 m/s just before it strikes a 44.0 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 41.4 m/s. Find the speed of the golf ball just after impact.

My attempt:
E=mv^2
E=(.210)(52.7)^2
E=583.2309J
Energy in the club after striking the ball
E=.210(41.4)^2
E=359.9316
The energy that is transferred to the golf ball = 583.2309-358.9316=223.2993J
Therefore the velocity must be 223.2993=.01v^2
v^2=22329.93
v=149.432
But this is wrong. What am i doing wrong?
 

Answers and Replies

  • #2
Your approach uses energy conservation. Energy might not be conserved in this collision (maybe there's some distortion of the ball or heating). What would be conserved, even if energy was not? Start with that.
 
  • #3
Matterwave
Science Advisor
Gold Member
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326
1) Can we assume perfectly elastic collision? If not, the energy is not conserved as some of the energy goes into deforming the ball and club.

2) Is the golfer still applying a torque on the golf club as he swings through? I would guess not if that information is not provided.

3) E=.5mv^2 not mv^2 , although that factor should cancel.

4) The ball is 44g not 10g as you have it.
 
  • #4
565
2
i corrected those mistakes and my answer is 71.238 which is still wrong
 
  • #5
2,017
85
EDIT: I blame being tired last night as the reason I did not to detect the major problem with what you are doing, I suck :P

read what physics girl said again. Energy is not conserved but '............' is.

this gives you your answer.
 
Last edited:
  • #6
565
2
the energy doesn't just disappear does it?
 

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