Answer:
Velocity of golf ball≅ 12.87mph
Explanation:
Using the theory of conservative momentum to both the club head and the golf ball we have;
Qi1 + Qi2= Qf1 + Qf2
Qi1: initial momentum for the club head
Qf1: final momentum for the club head
Qi2: initial momentum for the golf ball
Qf2: final momentum for the golf ball
momentum (Q) = Mass x velocity
which means the sum of the momentums of both club head and golf ball has to be the same before and after they have collided.
using the Coeficient of restitution e= 0.83 allows us to know what kind of collision we are dealing with, which is a partially elastic collision since
0> e=0.83 >1.
Qi1= 190 x 94= 17,860 mph.g
Qf1= 190 x Vf1
Qi2= 46 x 0= 0 mph.g
Qf2= 46x Vf2
Using the value of e to determine Vf2 as the final velocity of the golf ball:
e= [tex]\frac{Vf2- Vf1}{Vi1 - Vi2}[/tex]
0.83= [tex]\frac{Vf2 - Vf1}{94 - 0}[/tex]
Vf1= (0.83 x 94)+ Vf2
Vf1= 78.02 + Vf2
Qi1 + Qi2= Qf1 + Qf2
17,860 + 0 = 190xVf1 + 46xVf2
17,860= 190x (78.02+Vf2) + 46xVf2
17,860= 14,823+ 190xVf2+46xVf2
Vf2≅ 12.87mph
