Feynman’s Lectures Exercises 6.9

Before firing, the total momentum is:

${\displaystyle (M+m_b)V_0=0}$

After firing one bullet, the platform moves with velocity $V_p$. By momentum conservation, total momentum remains zero:

${\displaystyle M{V}_p+m_b{V}_b=0}$

Solving for $V_p$:

${\displaystyle V_p=-\frac{m_b{V}_b}{M}=-0.005\text{ m/s}}$

A negative sign indicates the platform moves opposite to the bullet’s direction.

A bullet travels 5 m at 500 m/s, thus its travel time is:

${\displaystyle t=\frac{5}{500}=\frac{1}{100}\text{ s}}$

Since bullets are fired every $\frac{1}{10}\text{ s}$, only one bullet is airborne at a time.

Approach 1: Considering 10 bullets per second

Each bullet provides a velocity of $0.005$ m/s to the platform, then it flies for some time and then it stops. In total, the platform moves for due to 10 bullets per seconds:

${\displaystyle t_{\text{total}}=nt=10\times \frac{1}{100}s=\frac{1}{10}s}$

Thus, the effective average velocity is:

${\displaystyle V=0.005\times \frac{1}{10}\times 10=5\times 10^{-4}\text{ m/s}}$

Approach 2: Using total distance traveled

The distance traveled by the platform per bullet fired is:

${\displaystyle s_p=V_{p}t=0.005\times \frac{1}{100}=5\times 10^{-5}\text{ m}}$

With 10 bullets fired each second, the effective velocity is:

${\displaystyle V=5\times 10^{-5}\times 10=5\times 10^{-4}\text{ m/s}}$