Question

Question 12 of 19 Lions can run at speeds up to approximately 80.0 km/h. A hungry 109 kg lion running northward at top speed attacks and holds onto a 36.0 kg Thomson's gazelle running eastward at 78.0 km/h. Find the final speed vf of the lion‑gazelle system just after the lion attacks.

Answer 1

Answer:

The final velocity of the lion-gazelle system is approximately 62.321 kilometers per hour.

Explanation:

Let suppose that both lion and the Thomson's gazelle collide each other inelastically, the use of the Principle of Linear Momentum Conservation suffices to describe the entire phenomenon:

$m_{L}\cdot \vec v_{L} + m_{T}\cdot \vec v_{T} = (m_{L}+m_{T})\cdot \vec v$ (1)

Where:

$m_{L}$ - Mass of the lion, in kilograms.

$m_{T}$ - Mass of the Thomson's gazelle, in kilograms.

$\vec v_{L}$ - Initial velocity of the lion, in meters per second.

$\vec v_{T}$ - Initial velocity of the Thomson's gazelle, in meters per second.

$\vec v$ - Final velocity of the lion-gazelle system, in meters per second.

Let suppose that both northward velocity and eastward velocity are positive.

If we know that $m_{L} = 109\,kg$, $\vec v_{L} = (0, 22.222)\left[\frac{m}{s} \right]$, $m_{T} = 36\,kg$ and $\vec v_{T} = (21.667, 0)\,\left[\frac{m}{s} \right]$, then the final velocity of the lion-gazelle system is:

$109\cdot (0,22.222)+36\cdot (21.667,0) = 147\cdot (v_{x},v_{y})$

$(v_{x}, v_{y}) = (0, 16.478) + (5.306, 0) \,\left[\frac{m}{s} \right]$

$(v_{x}, v_{y}) = (5.306, 16.478)\,\left[\frac{m}{s} \right]$

$(v_{x}, v_{y}) = (19.102, 59.321)\,\left[\frac{km}{h} \right]$

And the final speed of the lion-gazelle system is calculated by the Pythagorean Theorem:

$v = \sqrt{19.102^{2}+59.321^{2}}\,\left[\frac{km}{h} \right]$

$v \approx 62.321\,\frac{km}{h}$

The final velocity of the lion-gazelle system is approximately 62.321 kilometers per hour.