a golf ball of mass 0.045 kg is hit off the tee at a speed of 35 m/s. the golf club was in contact with the…

Explanation:

Step1: Recall impulse - momentum theorem

The impulse $J$ imparted to an object is equal to the change in its momentum. Since the ball starts from rest, $J = \Delta p=mv - mu$, where $u = 0$ (initial velocity), $m$ is the mass of the ball and $v$ is the final velocity. So $J=mv$. Given $m = 0.045\ kg$ and $v = 35\ m/s$. $J=0.045\times35$

Step2: Calculate impulse value

$J = 1.575\ kg\cdot m/s$. Rounding to two - significant figures, $J\approx1.6\ kg\cdot m/s$.

Step3: Recall the definition of impulse for part B

The impulse $J$ is also given by $J=\bar{F}\Delta t$, where $\bar{F}$ is the average force and $\Delta t$ is the time of contact. We know $J$ from part A and $\Delta t=3.30\times 10^{-3}\ s$. We can solve for $\bar{F}$ using $\bar{F}=\frac{J}{\Delta t}$. $\bar{F}=\frac{1.575}{3.30\times 10^{-3}}$

Step4: Calculate average - force value

$\bar{F}=\frac{1.575}{3.30\times 10^{-3}}\approx477.27\ N$. Rounding to two - significant figures, $\bar{F}\approx480\ N$.

Answer:

Part A: Value: $1.6$ Units: $kg\cdot m/s$ Part B: Value: $480$ Units: $N$