M1 help :(

Can somebody help me .I'm losing the will to live .The mark scheme doesn't explain much

Can somebody help me .I'm losing the will to live .The mark scheme doesn't explain much -.-


Posted from TSR Mobile

Can somebody help me .I'm losing the will to live .The mark scheme doesn't explain much -.-


Posted from TSR Mobile

How far did you get? Do you understand that when they hit each other they 'coalesce', meaning they 'come together' and move (to the right), so you should get:

Momentum before = momentum after

(0.7*4)+(0.3*1)=(0.7+0.3)v

Then just rearrange and work out v.

Not sure about the rest, damn the other two parts seem hard! Is this Mechanics 1B? That's what i'm doing, this seems a bit harder.

Hi!

Firstly, what sadeqrahman was doing is right, however it does not take into account the speeds - they decelerate, so at impact P is only travelling at (v=4+((-0.4)*2)) 3.2m/s, and Q at (v=1+((-0.4)*2)) 0.2m/s. Other than that, it is basically right - just replace the values.

Secondly, for (ii)(a) you need to take the acceleration as your gradient and the initial velocity as your c. For both they are accelerating at a negative amount (deceleration) so you end up with:

(P) y = -0.4x + 4

(Q) y = -0.4x + 1

Plotting these should give you two graphs which tell you the velocities. One thing you need to be aware of, however is that the force is due to friction - so when they go to -ve velocities, just draw across - friction will not decelerate from 0m/s.

With these graphs you can work out the distance travelled by each. Just work out the area below the lines, or use s=ut+1/2at^2. For P you should get

s = (4*3)+(1/2)(-0.4)(3^2)
s = 12-(0.2*9)
s = 10.2m

and for Q, as it goes to 0 and is linear just take the average velocity ((1+0)/2), and multiply by the time (3 seconds) to get 1.5m. Add 10.2 and 1.5 to get the total distance between them of 11.7m

(Please Note: What I have said may be completely wrong - I am still doing GCSEs and have not properly started the course and this is just from preliminary readings)

Good luck!

Two Planes (ThatPerson2).

Somebody's got a bit too much time on their hands - lol You Genius.

Hi!

Firstly, what sadeqrahman was doing is right, however it does not take into account the speeds - they decelerate, so at impact P is only travelling at (v=4+((-0.4)*2)) 3.2m/s, and Q at (v=1+((-0.4)*2)) 0.2m/s. Other than that, it is basically right - just replace the values.

Secondly, for (ii)(a) you need to take the acceleration as your gradient and the initial velocity as your c. For both they are accelerating at a negative amount (deceleration) so you end up with:

(P) y = -0.4x + 4

(Q) y = -0.4x + 1

Plotting these should give you two graphs which tell you the velocities. One thing you need to be aware of, however is that the force is due to friction - so when they go to -ve velocities, just draw across - friction will not decelerate from 0m/s.

With these graphs you can work out the distance travelled by each. Just work out the area below the lines, or use s=ut+1/2at^2. For P you should get

s = (4*3)+(1/2)(-0.4)(3^2)
s = 12-(0.2*9)
s = 10.2m

and for Q, as it goes to 0 and is linear just take the average velocity ((1+0)/2), and multiply by the time (3 seconds) to get 1.5m. Add 10.2 and 1.5 to get the total distance between them of 11.7m

(Edit: On part b I just realised you would multiply it by 2.5 seconds, so it would be 1.25m instead, gaining 11.45m.)

(Please Note: What I have said may be completely wrong - I am still doing GCSEs and have not properly started the course and this is just from preliminary readings)

Good luck!

Two Planes (ThatPerson2).