How Does a Hawk's Collision with a Pigeon Alter Its Flight Path and Speed?

[JerG90]

Homework Statement

A 5.4kg hawk flying swoops down at 65 degrees to the horizontal at 18m/s. The hawk grabs a pigeon flying 14 m/s horizontally. The pidgeon weighs 1.2kg. After the collision, what is the speed of the hawk (with the pidgeon) and at what angle to the horizontal is it now flying?

Homework Equations

The equations used in this chapter were P=MV and MA1VA1+MB1VB1= MTVT

The Attempt at a Solution

I don't know how to solve this problem other than using sine and cosine formulas.

I multiplied the masses/velocities together to find kg*m/s and I found that the hawk had 97.2kg*m/s and the pigeon had 16.8kg*m/s

By using a^2=b^2+c^2 -2(b)(c)(cosA)

I found a resultant of 105.4kg*m/s. Dividing by the total mass (6.6kg) I found a total velocity of 15.97m/s.

Then, to find the angle, I used law of sines.

105.4/sin115 = 16.8/sinX == 8.3 degrees

Add 25 degrees to make up for the hawk's dive. So that ends up being 33.3degrees.

So...can I have help verifying this?

 

[denverdoc]

Let me muddle thru my own way which is to first resolve the momentum and then conserve it:

P(hawk,x)=5.4*cos(65)*18=41.08
P(hawk,y)=5.4*sin(65)*18=-88.09 (negative sign since below horizontal)
P(pig,x)=1.2*14=16.8
p(pig,y)=0



Total Momentum(x)=41.08+16.8=57.88
Total Momentum(y)=88.09

direction = arc tan(88.09/57.88)=-56.7 degrees Intuitively this looks good as it causes a slight flattening on the hawks dive.

The velocity can be computed from above, but since our sol'ns are in disagreement--I get the compliment of the angle of 33 deg, no point for now and may just be a matter of how you are defining the angle, ie you may be right depending.