Solved Problem on Impulse, Linear Momentum and Collisions

Solved Problem on Linear Momentum

A cannon of a pirate ship of the eighteenth century had a mass of 2000 pounds and fired 10-pound projectiles at a speed of 300 m/s. The cannon is in the horizontal position, and the speed of the projectile is constant inside. Determine the speed recoil of the cannon. Assume 1 pound equal to 450 grams.

Problem data:

Mass of cannon: M = 2000 lb;
Mass of projectile: m = 10 lb;
Speed of projectile: v = 300 m/s.

Solution

First, we must convert the masses of the cannon and the ball given in pounds (lb) for kilograms (kg) used in the International System of Units (SI)

\[ \begin{gather} M=2000\;\mathrm{\cancel{lb}}\times\frac{450\;\mathrm{\cancel g}}{1\;\mathrm{\cancel{lb}}}\times\frac{1\;\mathrm{kg}}{1000\;\mathrm{\cancel g}}=900\;\mathrm{kg}\\[10pt] m=10\;\mathrm{\cancel{lb}}\times\frac{450\;\mathrm{\cancel{g}}}{1\;\mathrm{\cancel{lb}}}\times\frac{1\;\mathrm{kg}}{1000\;\mathrm{\cancel g}}=4,5\;\mathrm{kg} \end{gather} \]

The forces in the system are, in the vertical direction, we have the weight, \( \vec{W} \), and the normal reaction force, \( \vec N \), that cancel each other. In the horizontal direction, the force that the cannon does in the projectile, \( \vec F \), has the same magnitude as the force that the projectile does in the cannon, \( -{\vec F} \), in the opposite direction (Figure 1). So the cannon-projectile system is isolated from external forces and we can apply the Law of Conservation of Momentum.

Figure 1

The momentum before the shooting should be equal to the momentum after the shooting

\[ \begin{gather} p_i=p_f \tag{I} \end{gather} \]

Figure 2

Initially, the cannon and the projectile are at rest (Figure 2-A). When the cannon has triggered this act in the projectile and produces a speed \( \vec{v} \) forward, the projectile reacts in the cannon and causes it to recoil with speed \( -{\vec V} \) (Figure 2-B). The momentum is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {p=m v} \end{gather} \]

the condition (I) can be written as

\[ \begin{gather} mv_{0}+MV_{0}=mv+MV\\[5pt] (4.5\;\mathrm{kg})\times 0+(900\;\mathrm{kg})\times 0=(4.5\;\mathrm{kg})\times\left(300\;\mathrm{\frac{m}{s}}\right)+(900\;\mathrm{kg})V\\[5pt] \left(1350\;\mathrm{\frac{kg.m}{s}}\right)+(900\;\mathrm{kg})V=0\\[5pt] V=-{\frac{1350\;\mathrm{\frac{\cancel{kg}.m}{s}}}{900\;\mathrm{\cancel{kg}}}} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {V=-1.5\;\text{m/s}} \end{gather} \]

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