Question

In a dark abandoned hall, there are a robber and a policeman standing against each other by the opposite sides of the hall. Having heard steps echoing in the darkness, the criminal points his gun horizontally on the unsuspecting cop and is about to shoot. What he doesn't know is that a potential difference U = 144 V was brought between the floor and the ceiling of the hall, thus creating a homogeneous field acting (vertically downwards) on the fired bullet, since the bullet acquires a charge q as it travels through the muzzle. Find the least charge-to-mass ratio q/m of the bullet such that it hits the floor before it reaches the policeman (m denote the mass of the bullet). The robber and the policemen are separated by a distance of D = 100 m, the ceiling is d = 8 m above the floor and the bullet is fired horizontally with muzzle velocity v0 = 400 m s−1. Assume that the robber holds his gun h = 2 m above the floor. (g = 10 m s−2)

Answer 1

3\ \mathrm{C/kg}

Answer 2

Step 1: Horizontal flight time
The bullet travels a horizontal distance D = 100 m with speed v₀ = 400 m/s.

$$t = \frac{D}{v_{0}} = \frac{100}{400} = 0.25\ \mathrm{s}.$$

Step 2: Electric field magnitude
Potential difference U = 144 V across distance d = 8 m downward gives

$$E = \frac{U}{d} = \frac{144}{8} = 18\ \mathrm{V/m}.$$

Step 3: Vertical motion under combined acceleration
Downward acceleration:

$$a = g + \frac{q}{m}\,E.$$

Initial height is 2 m. To hit the floor before t = 0.25 s, the vertical drop must satisfy

$$y = \tfrac12\,a\,t^2 \ge 2\ \mathrm{m}.$$

Thus

$$\tfrac12\bigl(g + \tfrac{q}{m}E\bigr)(0.25)^2 \ge 2 \;\Longrightarrow\; (g + \tfrac{q}{m}E) \ge \frac{4}{0.0625}=64\ \mathrm{m/s^2}.$$

Step 4: Solve for q/m

$$10 + \frac{q}{m}\,18 \;\ge\; 64 \quad\Longrightarrow\quad \frac{q}{m} \;\ge\; \frac{54}{18} = 3\ \mathrm{C/kg}.$$

Therefore, the minimum charge‑to‑mass ratio required is

$$\boxed{\frac{q}{m} = 3\ \mathrm{C/kg}}.$$