A 4.7 kg bird is flying 48 m above the ground at a speed of 2.9 m/s. Calculate t

Question

A 4.7 kg bird is flying 48 m above the ground at a speed of 2.9 m/s. Calculate the KE, PE, and momentum of the bird. KE = J PE = J P = kg middot m/s A closed system consists of 2 objects. Initially, object A has a momentum of 41 kg middot m/s north and object B has a momentum of 61 kg middot m/s south. The two contacts have a head-on collision. Afterwards, object A is observed to have a momentum of 39 kg m/s south. After the collision, that is object B's momentum? kg middot m/s East West Up Down North South Left Right No direction because object B has stopped

Explanation / Answer

a] Kinetic energy of the bird = K.E = (1/2) mv2 = (1/2)4.7(2.9)2 = 19.7635 J

Potential energy of the bird = P.E = mgh = 4.7 x 9.8 x 48 = 2210.88 J.

b] The total momentum before and after the collision should be conserved.

Total momentum along North - South direction before the collision = 61 - 41 = 20 kgm/s (South direction)

after collision = (39 - p) kgm/s

therefore, 20 = 39 - p

=> p = 19 kgm/s in the North direction.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.