Question

A 1.00-kg duck is flying horizontally in the +x-direction at 20.0 m/s when seized (held onto) by an 0.800-kg hawk diving down with speed 30.0 m/s. The haw is coming in from behind and makes an angle of

30o relative to the vertical (i.e., -60o relative to +x-axis). What is the velocity of the two birds just after contact?

Answer 1

Applying the conservation of linear momentum, the combined velocity of the duck and hawk after contact is determined by considering the initial linear momenta of the two birds in motion.

To find the velocity of the duck and hawk just after contact, we can use the principle of conservation of linear momentum. The law of conservation of linear momentum states that the total linear momentum of an isolated system remains constant if no external forces act on it. In this case, the system consists of the duck and the hawk.

The linear momentum (p) is given by the product of mass (m) and velocity (v):

`\[ p = m \cdot v \]`

The initial linear momentum of the system is the sum of the linear momenta of the duck and the hawk before the contact:

`\[ p_{\text{initial}} = m_{\text{duck}} \cdot v_{\text{duck, initial}} + m_{\text{hawk}} \cdot v_{\text{hawk, initial}} \]`

After the contact, the two birds move together, and their combined mass is
`\(m_{\text{duck}} + m_{\text{hawk}}\). Let \(v_{\text{final}}\)` be the velocity of the combined system just after contact:

`\[ p_{\text{final}} = (m_{\text{duck}} + m_{\text{hawk}}) \cdot v_{\text{final}} \]`

By applying the conservation of linear momentum:

`\[ p_{\text{initial}} = p_{\text{final}} \]`

Now, substitute the given masses and velocities into these equations and solve for
`\(v_{\text{final}}\)`.

The velocity will have both horizontal and vertical components. The horizontal component can be found using trigonometry, as the hawk is making an angle of 30 degrees relative to the vertical.