Suppose that you have a mass of 45.7 kg and are standing on frictionless roller skates.Someone then throws you a 2.50 kg mass with a velocity of 14.5 m/s and you catch it.What will be your resultant velocity?(0.75 m/s)

Question

anonymous · Answer

This is a conservation of momentum problem.  
Starting momentum of system = Final momentum of system 

Where the system is the skater and the mass thrown

$$ P_{skater start} + P_{mass start} = P_{skater final} + P_{mass final}$$

$$ m_{skater} v_{skater start} + m_{mass}v_{mass start} = m_{skater}v_{skater final} + m_{mass}v_{mass final}$$

Starting velocity of skater is 0 [m/s] = V skater start therefore:

$$ 0 + m_{mass}v_{mass start} = m_{skater}v_{skater final} + m_{mass}v_{mass final}$$

After the mass is thrown and caught by the skater both the mass and the skater travel at the same velocity.   V skater final = V mass final = V final

$$ m_{mass}v_{mass start} = m_{skater}v_{final} + m_{mass}v_{final}$$

$$ m_{mass}v_{mass start} = (m_{skater}+ m_{mass})v_{final}$$

Then plug in the numbers and solve for Vfinal ... are you ok with this point?

anonymous · Answer

I plugged:45.7(0)=(2.50)(14.5)=4.57v+2.50v is this correct?

anonymous · Answer

45.7*

anonymous · Answer

$$(2.5)(14.5) [kg* m/s] = (45.7 + 2.5) [kg] *V_{final}$$

$$...  V_{final} = 0.75 [m/s]$$

anonymous · Answer

You should expect the skater plus mass combination will move slower than the original mass because of the additional mass from the skater.