Question

Two equal-mass birds collide and become entangled, moving at 18.9 m/s at 25° north of east immediately after the collision. before the collision, bird 1 is heading north and bird 2 is heading east. find bird 2's speed v just before the collision.

v = ______ m/s

Answer 1

To find the speed of bird 2 just before the collision, we can use the law of conservation of momentum. Bird 1's velocity components are 0 m/s in the x-direction and 18.9 m/s in the y-direction. Bird 2's velocity components are 18.9 m/s in the x-direction and 0 m/s in the y-direction. Therefore, bird 2's speed just before the collision is 26.7 m/s.

Step-by-step explanation:

To find the speed of bird 2 just before the collision, we can use the law of conservation of momentum. Since bird 1 is heading north and bird 2 is heading east before the collision, their initial velocities can be decomposed into their x and y components. Bird 1's velocity components are 0 m/s in the x-direction and 18.9 m/s in the y-direction. Bird 2's velocity components are 18.9 m/s in the x-direction and 0 m/s in the y-direction.

The total momentum before the collision is given by the sum of the momenta of bird 1 and bird 2. In the x-direction, the momentum before the collision is:

• Mx = m1v1x + m2v2x
• Mx = (m1)(0 m/s) + (m2)(18.9 m/s)
• Mx = 0 + (18.9 kg)(m/s)
• Mx = 18.9 kg m/s

The total momentum before the collision in the y-direction is:

• My = m1v1y + m2v2y
• My = (m1)(18.9 m/s) + (m2)(0 m/s)
• My = (18.9 kg)(m/s) + 0
• My = 18.9 kg m/s

The total momentum before the collision is:

• M = sqrt(Mx2 + My2)
• M = sqrt((18.9 kg m/s)2 + (18.9 kg m/s)2)
• M ≈ 26.7 kg m/s

The speed of bird 2 just before the collision, v, is equal to the magnitude of M:

• v = M
• v = 26.7 kg m/s

Therefore, bird 2's speed just before the collision is 26.7 m/s.