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

To find the final speed of the lion-gazelle system immediately after the attack, we need to apply the conservation of momentum.

Step-by-step explanation:

To find the final speed of the lion-gazelle system immediately after the attack, we need to apply the conservation of momentum. Momentum of an object is given by multiplying its mass and velocity.

We can calculate the momentum of the lion-gazelle system before the attack and equate it to the momentum after the attack. Assuming the lion and gazelle move as a single system after the attack, we have:

(Mass of lion * Speed of lion) + (Mass of gazelle * Speed of gazelle) = (Total mass of lion and gazelle after attack * Final speed of lion-gazelle system)

Substituting the given values, we get:

(177 kg * 81.8 km/hr) + (32 kg * 59.0 km/hr) = (177 kg + 32 kg) * Final speed of lion-gazelle system

Solving the equation for the final speed, we find the final speed of the lion-gazelle system immediately after the attack.

Answer 2

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

Step-by-step 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) = (m_(L))/(m_(L)+m_(G))\cdot \vec v_(L) + (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[(km)/(h) \right]` and
`\vec v_(G) = 59.0\cdot i\,\left[(km)/(h) \right]`, the final velocity of the lion-gazelle system is:

`\vec v_(F) = (177\,kg)/(177\,kg+32\,kg)\cdot \left(81.8\cdot j\right)\,\left[(km)/(h) \right] + (32\,kg)/(177\,kg+32\,kg)\cdot \left(59.0\cdot i\right)\,\left[(km)/(h) \right]`

`\vec v_(F) = 9.033\cdot i + 69.276\cdot j\,\left[(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(km)/(h) \right)^(2)+\left(69.276(km)/(h) \right)^(2)}`

`\|\vec v_(F)\| \approx 69.862\,(km)/(h)`

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