Answer to Question #97698 in Physics for Brianna

Question

A pyrotechnical expert needs to fire a 25 kg projectile from a launching device that has a barrel length of 2 meters. The projectile will need to be launched horizontally 1 km in 5 seconds. Calculate the force needed to launch the projectile

Answer 1

Let us use notation: $L = 2 m$ - length of the barrel, $m = 23 kg$ - mass of the object, $S = 1km = 1000m$ - horizontal range of the projectile, $T = 5 s$ - time to cover the distance $S$.

Let us assume that the object is moving inside the barrel with constant acceleration, until at the end of the barrel it reaches the initial speed of the projectile $v_0$, needed to cover the distance $S$. The force needed to accelerate the object inside the barrel in that way is $F = \frac{\Delta p}{\Delta t} = \frac{m(v_2 - v_1)}{\Delta T} = \frac{m v_0}{\Delta t}$, since the object changes the momentum from to $mv_0$.

Since the horizontal distance, covered in time $T$ is $S = v_0 T$, the initial speed is $v_0 = \frac{S}{T}$.

For accelerated motion inside the barrel, the following equations hold: $L = \frac{a (\Delta t)^2}{2}$, $v_0 = a \Delta t$. Dividing the first equation by second, obtain $\frac{L}{v_0} = \frac{\Delta t}{2}$, from where $\Delta t = \frac{2 L}{v_0}$ - the time of the motion inside the barrel.

Hence, using formula for $\Delta t$ and $v_0$ , obtain $F = \frac{m v_0}{\Delta t} = m \frac{v_0^2}{2 L} = \frac{m}{2 L} \frac{S^2}{T^2} = 250kN$.