Calculating Force Constant in a Spring Collision

  • Thread starter Thread starter PhysicsOPhun
  • Start date Start date
AI Thread Summary
In a collision between two blocks, one moving and one at rest, the maximum compression of a spring occurs when both blocks move together at half the initial speed. The kinetic energy of the moving block before the collision is calculated, and the energy stored in the spring at maximum compression is considered. The difference in kinetic energy before and after the collision indicates energy transfer to the spring. The discussion highlights the relationship between kinetic energy and potential energy in this inelastic collision scenario. Understanding these concepts is crucial for calculating the spring's force constant.
PhysicsOPhun
Messages
14
Reaction score
0
Two blocks, each with a mass 0.40 {\rm kg}, can slide without friction on a horizontal surface. Initially, block 1 is in motion with a speed v = 1.2 {\rm m/s}; block 2 is at rest. When block 1 collides with block 2, a spring bumper on block 1 is compressed. Maximum compression of the spring occurs when the two blocks move with the same speed, v/2 = 0.60 {\rm m/s}.


If the maximum compression of the spring is 1.9 {\rm cm}, what is its force constant?



So, I just posted a question but this one seems just like it. I guess I don't understand what effect each force has on one another and where to begin calculating it. Should I be thinking about the work energy theorm? How would I set this problem up? Thanks in advance to any that can tackle this. This by the way is nothing close to what we have been learning but I guess they are trying to challenge us...and I'm very curious
 
Physics news on Phys.org
It's an inelastic collision since the spring absorbs some of the energy.

Block 1 will decelerate and block 2 will accelerate, and meanwhile some energy will be stored in spring. But, one is told that maximum deflection occurs when the speed of both blocks is v/2 (0.6 m/s).

So what is the KE of block 1 before the collision, and what is the KE of both blocks at v/2? What is the significance of the difference?
 
well the kinetic energy of both blocks moving at the same velocity is exactly half of the kinetic energy of the first block in motion. which means? i have the same problem and I'm still lost. i'll take a guess though. since both the velocity and the kinetic energy of the first block are halved, can i assume that potential energy of the first block is halved as well?
 
Last edited:
yes i can :D thanks for the help. i hope the original poster gets this
 
Sillybean...

Hey Sillybean, you have a message! :redface:
 
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