Answer:
65.2 m/s
Explanation:
We are given that
Mass of golf club=m=200 g=[tex]\frac{200}{1000}=0.2kg[/tex]
1 kg=1000 g
Mass of golf ball=m'=46 g=[tex]\frac{46}{1000}=0.046kg[/tex]
Initial velocity of golf club=[tex]v_i[/tex]=55m/s
Final velocity of golf club=[tex]v_f=40m/s[/tex]
Initial velocity of golf ball=0m/s
By conservation law of momentum
Total initial momentum=Final total momentum
[tex]mv_i+m'v'_i=mv_f+m'v'_f[/tex]
Substitute the values
[tex]0.2(55)+(0.046)(0)=0.2(40)+(0.046)v'_f[/tex]
[tex]11=8+0.046v'_f[/tex]
[tex]0.046v'_f=11-8=3[/tex]
[tex]v'_f=\frac{3}{0.046}=65.2m/s[/tex]
Hence, the speed of the golf ball just after impact=65.2 m/s
