Why Does Momentum Seem Unequal in a Ball-Wall Collision?

  • Thread starter Thread starter alcatras
  • Start date Start date
  • Tags Tags
    Momentum
AI Thread Summary
Momentum conservation does not apply to the ball alone during its collision with the wall because an external force from the wall alters the ball's momentum. The system must include the wall and the Earth, as the wall absorbs some of the ball's momentum and transfers it to the Earth. This results in a net change in momentum for the entire system, not just the ball. The Earth's mass is so large that any change in its velocity is imperceptible. Understanding these principles clarifies why momentum appears unequal in this scenario.
alcatras
Messages
3
Reaction score
0
:smile: HELP is NEEDED! :smile:

A 1kg ball hits the wall with V= 5m/s and moves in opposite direction with V =4m/s. P1=5kgm/sec and P2=-4kgm/sec. But P1 must be equal to P2 from conservation of momentum , isn't that true?

THANKS :rolleyes:
 
Physics news on Phys.org
alcatras said:
:smile: HELP is NEEDED! :smile:

A 1kg ball hits the wall with V= 5m/s and moves in opposite direction with V =4m/s. P1=5kgm/sec and P2=-4kgm/sec. But P1 must be equal to P2 from conservation of momentum , isn't that true?

THANKS :rolleyes:
NO. Conservation of Momentum does not apply to the ball (as a single entity) because the wall applied a force to the ball, thus changing the ball's momentum. Anytime a net EXTERNAL force is applied to an entity, that entity's momentum will change (and not be conserved).


~~
 
Last edited:
alcatras said:
A 1kg ball hits the wall with V= 5m/s and moves in opposite direction with V =4m/s. P1=5kgm/sec and P2=-4kgm/sec. But P1 must be equal to P2 from conservation of momentum , isn't that true?

When you're considering conservation of momentum, you have to consider the momentum of the whole system, not just the ball. When it collides with the wall, some momentum is imparted to the particles in the wall. The wall, however, is stuck to the earth, so momentum is then imparted to the earth! Can you use the equation for momentum to answer why wouldn't notice such a change in the Earth's velocity?
 
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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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