-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
m1 help needed asap!
- Thread starter mz23
- Start date
- Messages
- 2,884
- Reaction score
- 415
- Points
- 93
time and distance will be same
now try it or else ill post it in detail
now try it or else ill post it in detail
- Messages
- 2,884
- Reaction score
- 415
- Points
- 93
hmmm ok
- Messages
- 2,884
- Reaction score
- 415
- Points
- 93
i did it
there is a way!
there is a way!
- Messages
- 2,884
- Reaction score
- 415
- Points
- 93
i meant the distance of both
- Messages
- 726
- Reaction score
- 52
- Points
- 0
Its pretty simple. To things are moving under the principal of constant acceleration. So, we only consider the constant acceleration formula's. Next, we've been given the height as 10 m. So, the sum of the distance that the two particles have travelled during their motion will be 10 m . The only difference is that for one gravitational pull is in the opposite direction.
Form two equations using the appropriate distance formula. Sum them and equate them to 10.
Form two equations using the appropriate distance formula. Sum them and equate them to 10.
??]
- Messages
- 2,884
- Reaction score
- 415
- Points
- 93
yup its big!but looks good
- Messages
- 2,884
- Reaction score
- 415
- Points
- 93
how do u make it?
- Messages
- 2,884
- Reaction score
- 415
- Points
- 93
great job!
Okay i hope u have the graph from M.S in front of you here is how it is,
From t=0 sec to t=2 seconds, both particles A and B are accelerating in opposing directions. That is why velocity of B is taken +ve and velocity of A is taken -ve
B stops after 2 seconds
From t= 2 sec to t=2.5 sec, particle A is slowing down due to gravity and stops at t= 2.5 sec (where v=0 and graph cuts x axis). This is the maximum point of A motion.
From t=2.5 sec to t=3 sec, particle A starts to move down again in downward direction and string becomes taut. Here the gradient of graph will be similar to that of motion from lowest to highest point. but it's position is in the positive direction as we have taken all the velocities travelling upwards as +ve.
Hope this helps and feel free to ask any other questions.
From t=0 sec to t=2 seconds, both particles A and B are accelerating in opposing directions. That is why velocity of B is taken +ve and velocity of A is taken -ve
B stops after 2 seconds
From t= 2 sec to t=2.5 sec, particle A is slowing down due to gravity and stops at t= 2.5 sec (where v=0 and graph cuts x axis). This is the maximum point of A motion.
From t=2.5 sec to t=3 sec, particle A starts to move down again in downward direction and string becomes taut. Here the gradient of graph will be similar to that of motion from lowest to highest point. but it's position is in the positive direction as we have taken all the velocities travelling upwards as +ve.
Hope this helps and feel free to ask any other questions.
thankyou!!! that helped alot!!! ok one more question...in the questions related to energy, what equation should we keep in mind as in loss in g.p.e=gain in k.e + WD against friction etc ....? and do you have any tips and tricks for m1 and specially questions related to energy?
Well im still not confident with my own energy and power concepts. However i have found the use of following helpful,
Always use the law of conservation of energy, which means;
Meachanical energy at start= Mechanical enrgy at end
=> where mechanical energy is the sum of kinetic and potential energies (which maybe loss or gain) and also the work done against other forces.
Just balance out all the energies around the equal to sign from the initial condition in the question at left hand side to the final condition given in the question at right hand side, and find out the uknown. That is what i have tried and is helpful for me. Just try to keep in mind every aspect of the question when making equations and balance it out.
Always use the law of conservation of energy, which means;
Meachanical energy at start= Mechanical enrgy at end
=> where mechanical energy is the sum of kinetic and potential energies (which maybe loss or gain) and also the work done against other forces.
Just balance out all the energies around the equal to sign from the initial condition in the question at left hand side to the final condition given in the question at right hand side, and find out the uknown. That is what i have tried and is helpful for me. Just try to keep in mind every aspect of the question when making equations and balance it out.