physics
posted by jillian on .
A 1.00kg duck is flying overhead at 1.50 m/s when a hunter fires straight up. The 0.010 0kg bullet is moving 100 m/s when it hits the duck and stays lodged in the duck's body. What is the speed of the duck and bullet immediately after the hit?

m1= 1kg
v1= 1.50 m/s
=====
m2= 0.010kg
v2= 100m/s
=====
m1*v1= 1kg*1.50kg= 1.50kgm/s
m2*v2= 0.010kg*100m/s= 1.0kgm/s
=====
mTotal= 1.010kg
Since the bullet is shot in a vertical movement, and the duck was flying horizontally, then we can use Pythagorean to solve for the final velocity(vF).
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Velocity in the xcomponent(vx)
> (mT)(vx)= momentum of duck.
> (1.010kg((vx)= 1.50kgm/s
> (vx)= (1.50kgm/s)/(1.010kgm/s)
> (vx)= 1.485m/s
....
Velocity in the ycomponent (vy)
> (mT)(vy)= 1.0kgm/s
> (1.010kg)(vy)= (1.0kgm/s)
> (vy)= (1.0kgm/s)/(1.010kgm/s)
> (vy)= (0.99009m/s)
=====
*Now we can use Pythagorean to solve for the final velocity!
> a^2 + b^2 = c^2
> (1.485m/s)^2 + (0.99009m/s)^2 = c^2
> 2.2055m/s + 0.9818m/s = c^2
> 3.1871m/s = c^2
> c = √(3.1871m/s)
> c = 1.785 m/s!
* Your final velocity is 1.785m/s!