Question

1. in a typical golf swing, the club is in contact with the ball for about 0.010 s. if the 45 g ball acquires a speed of 65 m/s, the magnitude of the force exerted by the club on the ball is (a) 1000 n (b) 45 n (c) 292.5 n (d) 5000 n (e) none of the above

Explanation:

Step1: Identify mass and convert units

The mass of the ball $m = 45\ g=0.045\ kg$.

Step2: Identify time and velocity

The time of contact $\Delta t = 0.010\ s$, and the final - velocity of the ball $v = 65\ m/s$ (initial velocity $u = 0\ m/s$).

Step3: Use impulse - momentum theorem

The impulse - momentum theorem states that $F\Delta t=\Delta p=m(v - u)$. Since $u = 0\ m/s$, we have $F=\frac{mv}{\Delta t}$.
Substitute $m = 0.045\ kg$, $v = 65\ m/s$, and $\Delta t=0.010\ s$ into the formula:
$F=\frac{0.045\times65}{0.010}=\frac{2.925}{0.010}=292.5\ N$.

Answer:

C. 292.5 N