stuck on dynamics question (1 Viewer)

[waltssillyhat]

i found part a, which was 40J, but i don't know how to use the information given to find part b and c. i assumed that the trolley came to rest at 5cm, since that was when the force-compression graph broke off, and my answers for b and c were correct, however i don't think my working was correct, as i didn't use all the info given in the question (and i basically just assumed when it would come to rest). pls help with b and c!!!

 

[cheesynooby]

i found part a, which was 40J, but i don't know how to use the information given to find part b and c. i assumed that the trolley came to rest at 5cm, since that was when the force-compression graph broke off, and my answers for b and c were correct, however i don't think my working was correct, as i didn't use all the info given in the question (and i basically just assumed when it would come to rest). pls help with b and c!!!
View attachment 47582View attachment 47583

I believe you should assume the kinetic energy is fully converted to elastic potential energy once the trolley is at rest, which is then converted back into the trolley's kinetic energy at the end.
So the final velocity of the trolley will give you the kinetic energy and elastic potential energy at rest.
That can then be used to find the compression for (c) by setting the area under the graph (representing work) = the elastic potential energy at rest. (which should be 5cm)

 

[waltssillyhat]

I believe you should assume the kinetic energy is fully converted to elastic potential energy once the trolley is at rest, which is then converted back into the trolley's kinetic energy at the end.
So the final velocity of the trolley will give you the kinetic energy and elastic potential energy at rest.
That can then be used to find the compression for (c) by setting the area under the graph (representing work) = the elastic potential energy at rest. (which should be 5cm)

ahhh thank you so much!!! i was able to work it out <3

 

[wizzkids]

This qualifies as an elastic "collision", so treat the problem as conservation of momentum and conservation of energy. I won't show any working here, I will just outline the best solution.
(a) energy is force x distance. The elastic potential energy E_p stored in the spring is equal to the area under the force vs. distance graph.
(b) The system will have the same total energy E_p + E_k before the collision, during the collision and after the collision.
(c) using the results from (a) and (b) find the value of compression x that gives E_p = initial E_k
(d) you should discuss the force exerted by the spring to the left which gives a change of momentum (impulse) to the left.
(e) The material property of the spring is elastic torsion, during which no energy is lost, i.e. energy is conserved. The macroscopic property of the spring is Hooke's Law and the elastic spring constant "k" defined by F= -k Δx. You should calculate k from the slope of the graph provided in order to give an extensive answer.

 

[waltssillyhat]

This qualifies as an elastic "collision", so treat the problem as conservation of momentum and conservation of energy. I won't show any working here, I will just outline the best solution.
(a) energy is force x distance. The elastic potential energy E_p stored in the spring is equal to the area under the force vs. distance graph.
(b) The system will have the same total energy E_p + E_k before the collision, during the collision and after the collision.
(c) using the results from (a) and (b) find the value of compression x that gives E_p = initial E_k
(d) you should discuss the force exerted by the spring to the left which gives a change of momentum (impulse) to the left.
(e) The material property of the spring is elastic torsion, during which no energy is lost, i.e. energy is conserved. The macroscopic property of the spring is Hooke's Law and the elastic spring constant "k" defined by F= -k Δx. You should calculate k from the slope of the graph provided in order to give an extensive answer.

thank u for the help!!