physics momentum and impulse help PLZZZ test tomorrow!!!?

  • Thread starter Thread starter Devi T
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Devi T

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A 59 kg physics student is riding her 220 kg Harley at 12 m/s when she has a head-on collision with a 2.1 kg pigeon flying the opposite direction at 44 m/s. The bird is still on the motorcycle after the collision. How fast is the motorcycle going after the collision?

can you please show your work soi know how to do it
cuz i have a test tomorrow and i need to know how to do it
thanks in advance :]
 
Conservation of momentum.

Before collision, momentum = 220 kg*12 m/sec - 2.1 kg * 44 m/sec
After collision, momentum = (220 + 2.1) * v and it's the same as the momentum before.

220*12 - 2.1*44 = (220 + 2.1)*v

The thing you need to remember is that unless there are external forces such as friction (which you probably won't have), momentum is always conserved, always the same before and after the collision. This is true whether the collision is elastic or inelastic. Your question, where two objects stick together, is an inelastic collision.

Kinetic energy is conserved if the collision is elastic. In that case you can also write the expression for KE before the collision and the KE after.
 
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