A baseball is thrown from the outfield toward the home plate

AI Thread Summary
The discussion centers on the conservation of momentum in a baseball thrown from the outfield toward home plate, focusing on two key questions. In part A, it is concluded that momentum is not conserved for the ball alone due to the influence of external forces, specifically gravity. In part B, the momentum of the earth-ball system is considered conserved, as the forces acting between the ball and the Earth are internal, thus maintaining a constant total momentum. The conversation also touches on the complexities of defining the system boundaries, particularly regarding air resistance and whether the atmosphere is included. Overall, the analysis emphasizes the distinction between isolated and non-isolated systems in the context of momentum conservation.
Apple
Messages
3
Reaction score
0
1. The problem statement, all variables, and given/known data
A baseball is thrown from the outfield toward the home plate. a) Neglecting air resistance, is the momentum of the ball conserved during its flight? explain b) Neglecting the air resistance, is the momentum of the earth-ball system conserved during baseball’s flight? explain.

Homework Equations



3. The Attempt at a Solution [/B]
I think the statement in part A is false and part B is true, I just don't know if I'm am right please explain the answers. Thank in advance!
 
Physics news on Phys.org
Apple said:
1. The problem statement, all variables, and given/known data
A baseball is thrown from the outfield toward the home plate. a) Neglecting air resistance, is the momentum of the ball conserved during its flight? explain b) Neglecting the air resistance, is the momentum of the earth-ball system conserved during baseball’s flight? explain.

Homework Equations



3. The Attempt at a Solution [/B]
I think the statement in part A is false and part B is true, I just don't know if I'm am right please explain the answers. Thank in advance!

We are not allowed to "explain the answers"; we are just allowed to offer hints. Surely you know why you gave those answers. What were your reasons?
 
  • Like
Likes Apple
Ray Vickson said:
We are not allowed to "explain the answers"; we are just allowed to offer hints. Surely you know why you gave those answers. What were your reasons?
I assumed that there were no external forces since air resistance was neglected. So, when no external forces act on an object the total momentum is constant. The Earth in the system causes the external force (mg) to become internal to the system and is canceled out by the third law force pair of the gravitational force of the ball on the earth.Then there would be no more outside force that would be able to change the kinetic energy.
 
Apple said:
I assumed that there were no external forces since air resistance was neglected. So, when no external forces act on an object the total momentum is constant. The Earth in the system causes the external force (mg) to become internal to the system and is canceled out by the third law force pair of the gravitational force of the ball on the earth.Then there would be no more outside force that would be able to change the kinetic energy.

I can't actually make out from that what your answers are. Try, instead:

In a) momentum is conserved/not conserved [delete as applicable] because ...
 
  • Like
Likes Apple
PeroK said:
I can't actually make out from that what your answers are. Try, instead:

In a) momentum is conserved/not conserved [delete as applicable] because ...
In a) momentum is not conserved because if there is an outside force acting on the ball (Gravity is an external force).
In b) momentum is conserved when the earth-ball system has no net momentum its final and initial momentum is the same, therefore it is constant.
 
Apple said:
In a) momentum is not conserved because if there is an outside force acting on the ball (Gravity is an external force).
In b) momentum is conserved when the earth-ball system has no net momentum its final and initial momentum is the same, therefore it is constant.

I think you've got the idea, although for b) you seem to have expressed things in an odd way. Did you think about using the term "internal forces" or "Newton's third law"?

That said, question b) is a bit odd, in that it mentions air resistance. That seems odd to me, because if you take the atmosphere out of the system, then the (solid) Earth and the ball are no longer a closed system. But, if the air is part of the Earth, then you have a closed system.
 
  • Like
Likes Apple
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top