- #1
XanziBar
- 46
- 0
Homework Statement
In a typical tee shot, a golf ball is hit by the 300 g head of a club moving at a speed of 40 m/s. The collision with the ball happens so fast that the collision can be treated as the collision of a 300 g mass with a stationary ball—the shaft of the club and the golfer can be ignored. The 46 g ball takes off with a speed of 70 m/s.
A) What is the change in momentum of the ball?
B)What is the speed of the club head immediately after the collision?
C) What fraction of the club’s kinetic energy is transferred to the ball?
D) What fraction of the club’s kinetic energy is “lost” to thermal energy?
Homework Equations
Pi=Pf
KE=.5*m*v^2
P=mV
The Attempt at a Solution
For A I did Pf-Pi=.046*70-0=3.22 Ns
For B I did conservation of momentum: .3*40=.3*v+.046*70
v=(.3*40-.046*70)/.3=29.3 m/s
For C the initial KE of the club is .5*.3*40^2=240 J
the final KE of the ball is 0.5*0.046*70*70=112.7 J
112.7/240=.47
for D
KEi=240 J from before and the final is
0.5*0.3*29.3*29.3 + 0.5*0.046*70*70=241.5 J
which is where I am stuck because I am confused why the initial energy is less than the final energy. Help?
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