A hungry 177177 kg lion running northward at 75.575.5 km/hr attacks and holds onto a 32.332.3 kg Thomson's gazelle running eastward at 60.660.6 km/hr. Find the final speed of the lion–gazelle system immediately after the attack.

73.2 km/hr

Explanation:

This is a momentum conservation type of problem, here we need to use the fact that the initial momentum is equal to the final momentum.

Remember that the momentum is written as:

P = m*v

m = mass

v = velicity

At the beginning, we have:

A lion of 177 kg running at velocity of 75.5 km/hr

Then the momentum of the lion is P1 = 177 kg*75.5 km/hr= (13,363.5)kg*km/hr

The gazelle has a mass of 32.3 kg, and a velocity of 60.6 km/hr

Then the momentum of the gazelle is:

P2 = 32.3kg*60.6km/hr = 1,957.38 kg*km/hr

The total initial momentum is equal to P1 + P2, then we have:

P = 1,957.38 kg*km/hr + 13,363.5 kg*km/hr = 15,320.9 kg*km/hr

Now, after the lion catches the gazelle, we can think it as only one object moving with mass equal to the mass of the lion plus the mass of the gazelle and velocity V

Then the final momentum will be:

Pf = (177kg + 32.3kg)*V = 209.3kg*V

And this needs to be equal to the initial momentum, then:

209.3kg*V = 15,320.9 kg*km/hr

V = (15,320.9 kg*km/hr)/209.3kg = 73.2 km/hr

Then the final speed of the lion-gazelle system is 73.2 km/hr

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