Understanding Momentum and Impulse: A Baseball Problem

  • Thread starter Thread starter husky88
  • Start date Start date
AI Thread Summary
The discussion focuses on a physics problem involving momentum and impulse related to a baseball's change in direction. The user correctly calculated the change in momentum as 24 kg*m/s but is confused about the direction, interpreting it as 22.5° East of South, while the answer key states 135°, indicating a different reference frame. The user also calculated the impulse as 24 N*s at the same angle but received a different direction from the answer key, which specifies 22.5° West of North. The confusion stems from the interpretation of vector subtraction and the reference angles used in the problem. Clarification on vector direction and the resultant vector's orientation is needed to resolve these discrepancies.
husky88
Messages
79
Reaction score
0
Hi. I was wondering if anyone could help me with this problem.

A baseball of mass 850 g traveling at 15 m/s due south is caught and thrown northwest at the same speed during a time interval of 0.85 s.
(a) What is the magnitude of the change in the momentum of the ball during this time interval?
(b) What is the direction of the change in the ball's momentum during this time interval?
(c) What are the magnitude and the direction of the impulse delivered to the ball during this time interval?

There are more questions, but I know how to do those.
I have solved the problem, but I got different answers from the answer key. I don't know what I am missing.

(a) I got this one correct 24 kg*m/s.
(b) I got 22.5° East of South. They say it is 135°, no direction specified. 135° is the angle between the two vectors. Why would that be the direction of the change in the momentum?
(c) I got 24 N*s 22.5° East of South. They say it is 24N*s 22.5° West of North. Why is the direction of the change in momentum different from the direction of the impulse? Why do I have a direction of East of South and they say it is West of North?

Any help would be greatly appreciated. Thank you.
 
Physics news on Phys.org
b) My interpretation of "change" is "new vector minus old vector" (if my weight changes from x to x+20, the change is +20, not -20), so I get 22.5° west of north.
 
Thank you so much. :smile:
Subtraction of vectors is very unclear to me. As to which way the resultant vector points, but I think I got it.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top