Question

A hungry 177 kg lion running northward at 81.8 km/hr attacks and holds onto a 32.0 kg Thomson's gazelle running eastward at 59.0 km/hr. Find the final speed of the lion–gazelle system immediately after the attack.

Answer 1

Answer:

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

Explanation:

Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:

$\vec p_{L} + \vec p_{G} = \vec p_{F}$

Where:

$\vec p_{L}$ - Linear momentum of the lion, measured in kilograms-meters per second.

$\vec p_{G}$ - Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.

$\vec p_{F}$ - Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.

After using the definition of momentum, the system is expanded:

$m_{L}\cdot \vec v_{L} + m_{G}\cdot \vec v_{G} = (m_{L} + m_{G})\cdot \vec v_{F}$

Vectorially speaking, the final velocity of the lion-gazelle system is:

$\vec v_{F} = \frac{m_{L}}{m_{L}+m_{G}}\cdot \vec v_{L} + \frac{m_{G}}{m_{L}+m_{G}}\cdot \vec v_{G}$

Where:

$m_{L}$, $m_{G}$ - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.

$\vec v_{L}$, $\vec v_{G}$, $\vec v_{F}$ - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.

If $m_{L} = 177\,kg$, $m_{G} = 32\,kg$, $\vec v_{L} = 81.8\cdot j\,\left[\frac{km}{h} \right]$ and $\vec v_{G} = 59.0\cdot i\,\left[\frac{km}{h} \right]$, the final velocity of the lion-gazelle system is:

$\vec v_{F} = \frac{177\,kg}{177\,kg+32\,kg}\cdot \left(81.8\cdot j\right)\,\left[\frac{km}{h} \right] + \frac{32\,kg}{177\,kg+32\,kg}\cdot \left(59.0\cdot i\right)\,\left[\frac{km}{h} \right]$

$\vec v_{F} = 9.033\cdot i + 69.276\cdot j\,\left[\frac{km}{h} \right]$

The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:

$\|\vec v_{F}\| = \sqrt{\left(9.033\frac{km}{h} \right)^{2}+\left(69.276\frac{km}{h} \right)^{2}}$

$\|\vec v_{F}\| \approx 69.862\,\frac{km}{h}$

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.