(practice exam question) Flying in horizontally, a 1.00 kg cockatoo lands on a h
ID: 2025260 • Letter: #
Question
(practice exam question)
Flying in horizontally, a 1.00 kg cockatoo lands on a hanging bird feeder table. The table weighs 4.5 kg and swings freely from a 3.5 m rope. After the landing, the bird and table together are observed to swing to a maximum height of 20 cm relative to the stationary position. When answering the questions below, be sure to state any physical laws and assumptions you make use of. What was the cockatoo's initial velocity? (ii) Determine if the landing was an elastic or inelastic collision.Explanation / Answer
When the bird+table whose mass is 5.5 kg swung 0.20 m higher, they gained 10.791 J of gravitational potential energy, which means that is what their kinetic energy was immediately after collision. That means the velocity for bird+table immediately after collision was [solving for v in (1/2)mv2] v = (2K/m)1/2 = 1.981 m/s.
Therefore their momentum immediately after the collision was mv = 10.895 kg-m/s. Now momentum is always conserved, elastic or inelastic, so that had to be the incoming momentum of the bird. There the bird's velocity was:
10.895 m/s [just divide out its mass of 1.0 kg to get that.]
To see whether or not the collision was elastic or not, compare the incoming kinetic energy to the 10.791 J of kinetic energy immediately after the collision. But it is obvious without doing the calculation:
(1/2)(1.0 kg)(10.895 m/s)2 is much more than 10.791, so the collision was inelastic.
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