# Collision problem (w/ angles)

• 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?

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.

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