Answer to Question #110793 in Classical Mechanics for Jazz Harris-Urlich

Question

A reckless hunter is on a boat that is stationary in the middle of a lake.

They fire their gun horizontally 10 times. Each time they fire, a 12 g bullet leaves the gun at a speed of 410 m s−1.

What is the speed of the boat after they have finished firing their gun if the total mass of the boat, hunter, and gun is 150 kg? (in m s−1 to 2 s.f)

Assume no drag forces from the water.

Answer 1

Each time the hunter fires, the mass of the boat decreases for 12 g. After the 10th shot, the change in the total mass of the hunter, boat, and gun will be of order

$$\frac{0.012\cdot10}{150}=0.0008.$$

Therefore, we neglect the fact that the mass decreases because it will not influence the required precision.

To find the final speed of the boat, apply momentum conservation principle.

During the first shot we have: zero initial momentum of the system, final momentum is composed of a bullet $mv$ and the boat $mu_1$:

$$0=mv-Mu_1,\\ u_1=v\frac{m}{M}=0.033\text{ m/s}.$$

Second shot. Initial momentum: boat moving with speed $u_1$. Final momentum: boat at speed $u_2$ and a bullet at $v$:

$$-Mu_1=m(v-u_1)-Mu_2,\\ u_2=u_1+\frac{m}{M}(v-u_1)=0.066\text{ m/s}.$$

Third shot. Initial momentum: boat moving with speed $u_2$. Final momentum: boat at speed​ $u_3$ and a bullet at $v$:

$$-Mu_2=m(v-u_2)-Mu_3,\\ u_3=u_2+\frac{m}{M}(v-u_2)=0.098\text{ m/s}.\\$$

If we continue this way, at the 10th shot the speed of the boat will be

$$u_4=u_3+(m/M)(v-u_3)=0.13\text{ m/s},\\ u_5=0.16\text{ m/s},\\ u_6=0.19\text{ m/s},\\ u_7=0.23\text{ m/s},\\ u_8=0.26\text{ m/s},\\ u_9=0.29\text{ m/s},\\ u_{10}=0.32\text{ m/s}.\\$$