Posted by jillian on Monday, March 29, 2010 at 5:00pm.
A 1.00-kg duck is flying overhead at 1.50 m/s when a hunter fires straight up. The 0.010 0-kg 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?
- physics - Lee, Thursday, March 28, 2013 at 1:50am
v1= 1.50 m/s
m1*v1= 1kg*1.50kg= 1.50kg-m/s
m2*v2= 0.010kg*100m/s= 1.0kg-m/s
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).
Velocity in the x-component(vx)
-> (mT)(vx)= momentum of duck.
--> (1.010kg((vx)= 1.50kg-m/s
---> (vx)= (1.50kg-m/s)/(1.010kg-m/s)
----> (vx)= 1.485m/s
Velocity in the y-component (vy)
-> (mT)(vy)= 1.0kg-m/s
--> (1.010kg)(vy)= (1.0kg-m/s)
---> (vy)= (1.0kg-m/s)/(1.010kg-m/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!
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