## anonymous one year ago A canoeist on a lake throws a 0.145 kg baseball at 40.0 m/s to a friend in a different canoe. Each canoe with its contents has a mass of 120 kg, and is at rest initially. How fast is the first canoe moving after the pitch, assuming no friction with the water? A. 0.048 m/s B. 0.33 m/s C. 20 m/s D. 40 m/s ** :/

1. Michele_Laino

here, since we can neglet any friction forces acting on our canoe, we have to apply the theorem of conservation of total momentum

2. anonymous

ok!

3. Michele_Laino

here is the situation described in your problem: |dw:1433009286262:dw|

4. anonymous

ok!

5. Michele_Laino

the momentum of the ball is: $\Large mv = 0.145 \times 40 = ...$

6. Michele_Laino

the speed of the ball is referred to the water, and the water is at rest with respect to the earth

7. anonymous

so 5.8?

8. anonymous

5.8 = mv?

9. Michele_Laino

ok! that's right!

10. anonymous

yay! what happens next?

11. Michele_Laino

the momentum gained, by the system canoeist+canoe is: $\Large MV = 120 \times V$

12. Michele_Laino

now both of those momentum have to be equal each other, so we can write: $\Large 120 \times V = 5.8$

13. anonymous

so 5.8=120*V V=0.0483333...? so our solution is choice A?

14. Michele_Laino

correct!

15. anonymous

yay!! thank you:)

16. Michele_Laino

thanks!! :)