How Does Adding a Spring Affect Collision Outcomes?

AI Thread Summary
Adding a spring in front of mass B alters the collision dynamics between mass A and mass B. In the first scenario, mass A collides with mass B, transferring momentum and resulting in mass B moving, while mass A may remain stationary if they are of equal mass. In the second scenario, the spring absorbs some energy during the collision, which affects the distance mass B travels compared to the first scenario. If the spring is sufficiently elastic, it can either propel mass A away or allow it to roll away, depending on its mass. Overall, while momentum is conserved in both scenarios, the presence of the spring changes the energy transfer and final velocities.
chandran
Messages
137
Reaction score
1
i would like to discuss the following two scenarios

Scenario 1)Mass A(velocity>0) moves towards Mass B(velocity=) and hits it. The final velocity is derived from applying the momentum conservation principle.

2)Scenario 2) The scenario is same as in scenario 1) but a spring is included in front of mass B.


The above two problems are given in a physics web.

I want to understand what is the difference between 1) and 2)
 
Physics news on Phys.org
Although i am not possitive, i think this may help.

Scenerio A: Object A strikes Object B and energy is transfered. As a result, Object A transfers momentum to Object B, and Object B then moves.

Scenerio B: Object A strikes the spring, which in turn exerts force of Object B. When the spring is hit, some of the energy is absorbed by the spring, resulting in Object B moving, but going a shorter distance than in scenerio A.

In Scenerio A, Object A should remain not move after coming into contact with Object B (if they are the same size.)

In Scenerio B, depending on the springyness of the spring, Object A would either be shot away as a result of the reciprication produced by the spring. If not shot away, Object A may roll away depending on its mass.

Hope this helps!
 
chandran said:
Scenario 1)Mass A(velocity>0) moves towards Mass B(velocity=) and hits it. The final velocity is derived from applying the momentum conservation principle.
If the two masses stick together after colliding (you didn't specify), this is an example of an inelastic collision. KE is not conserved.

2)Scenario 2) The scenario is same as in scenario 1) but a spring is included in front of mass B.
Assuming no energy is lost in the spring, this would be an example of an elastic collision. KE of the masses is conserved (assuming the spring's mass can be neglected).

In both collisions momentum is conserved.
 
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 '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