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

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

Step-by-step 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[(m)/(s) \right]`,
`m_(T) = 36\,kg` and
`\vec v_(T) = (21.667, 0)\,\left[(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[(m)/(s) \right]`

`(v_(x), v_(y)) = (5.306, 16.478)\,\left[(m)/(s) \right]`

`(v_(x), v_(y)) = (19.102, 59.321)\,\left[(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[(km)/(h) \right]`

`v \approx 62.321\,(km)/(h)`

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