Answer
a. 90 $\frac{kg \cdot m}{s}$
b. 11000N
Work Step by Step
a. The impulse given to the nail is the opposite of the hammer’s change in momentum. Let the positive direction be the direction of the hammer’s motion.
$$-\Delta p_{hammer} = -m \Delta v = -(12 kg)(0-7.5 \frac{m}{s}) = 90 \frac{kg \cdot m}{s}$$
b. Find the average force by dividing the impulse by the time the objects are in contact.
$$F_{av}=\frac{\Delta p}{\Delta t}=\frac{90 kg \cdot m/s}{8.0 \times 10^{-3} s}\approx 11000 N$$
