Mechanics M1: Post your doubt here

[AhShun]

Thank you !
Can you explain me for no, 4 & 5(ii) also ?
I can't figure it out
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_43.pdf

Nov10 p43 no5(ii)

 

[bamteck]

Nov10 p43 no4

I can't understand

 

[AhShun]

I can't understand

which part (i) or (ii)?

 

[bamteck]

which part (i) or (ii)?

no. 4 both part

 

[AhShun]

when i first saw the question,it was very confusing
i drew the graph first
it says 0<t<4 it has an acceleration= +0.75(therefore the gradient of the graph must be positive)==>straight line going up
4<t<54 acceleration= 0(gradient of graph=0)
54<t<60 acceleration= -0.5(gradient negative)==> straight line goes down

I do not know the speed, I put it 'u'

0<t<4
acceleration=gradient
0.75 =(u-0)/(4-0)
u=0.75 x 4
=3

 

[AhShun]

no. 4 both part

distance=area under graph for velocity-time graph
=area triangle for 1st 4s + area of rectangle(4<t<54) + area triangle(54<t<60)

If you still don't understand,don't hesitate to ask again

 

[bamteck]

when i first saw the question,it was very confusing
i drew the graph first
it says 0<t<4 it has an acceleration= +0.75(therefore the gradient of the graph must be positive)==>straight line going up
4<t<54 acceleration= 0(gradient of graph=0)
54<t<60 acceleration= -0.5(gradient negative)==> straight line goes down

I do not know the speed, I put it 'u'

0<t<4
acceleration=gradient
0.75 =(u-0)/(4-0)
u=0.75 x 4
=3

Yeah, now its clear for me !
Thank you mate You are very nice

 

[GorgeousEyes]

A body will have greatest speed when its kinetic energy is highest and kinetic energy will be highest when P.E is lowest. So use mgh = 0.5mv^2. And put the lowest possible value of h i.e 2.45m. For the ii.) part use K.E = P.E lost but use h = 2.45 - 1.2 while calculating P.E because it is the actual height of N from L. Again for iii.) use the K.E that you calculated in ii.) to find out least value of v using the equation k.E = 0.5mv^2. Now I guess you ended up with all correct values.

Please hit the "Like" button if you think I helped you.

Why Lowest value of H is 2.45 ? isn't it 1.2 ?

 

[GorgeousEyes]

Jun08 no. 1 ii
no.4 iii

 

[Mr Me]

Why Lowest value of H is 2.45 ? isn't it 1.2 ?

The lowest value of H is 2.45 at point M. Look at the left hand side of the figure and you will understand it.

 

[Mr Me]

Jun08 no. 1 ii
no.4 iii

For 1 ii.) Use a = gSinα and replace the values of a and g then you will get the value of α.
For 4. iii.) Use the equation mgh = 0.5mv^2 and replace the values of m, and v (i.e v = 8 ) and perform the necessary calculations. Then you will get the actual value of h.

 

[GorgeousEyes]

The lowest value of H is 2.45 at point M. Look at the left hand side of the figure and you will understand it.

Still can't get it

 

[bamteck]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_41.pdf

Please someone help me for no. 5 (ii) (iii)

 

[GorgeousEyes]

Nov08 7 , ii please

 

[GorgeousEyes]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_41.pdf

Please someone help me for no. 5 (ii) (iii)

Ia it June 2011 , 41 ?

 

[Asheeta]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_41.pdf

Please someone help me for no. 5 (ii) (iii)

so in the first part you've already found V i.e the speed at t=600 s and it is 15m/s and T=3600 s
(ii)
so you'll find the sketch attached. AB= (1/2)*(2600+3600)*15 =46500 m
(iii)
from the pic you can see that at v=7.5 m/s there are 2 values of t, t1 and t2.
using (0,0) (t1,7.5) m=0.025, find t1, t1= 300 s
using (3200,15) (t2,7.5) m = -0.0375 find t2, t2 = 3400 s
Hope it helped

 

[bamteck]

Ia it June 2011 , 41 ?

Yeah !

 

[bamteck]

so in the first part you've already found V i.e the speed at t=600 s and it is 15m/s and T=3600 s
(ii)
so you'll find the sketch attached. AB= (1/2)*(2600+3600)*15 =46500 m
(iii)
from the pic you can see that at v=7.5 m/s there are 2 values of t, t1 and t2.
using (0,0) (t1,7.5) m=0.025, find t1, t1= 300 s
using (3200,15) (t2,7.5) m = -0.0375 find t2, t2 = 3400 s
Hope it helped

Thanks Aheeta

 

[GorgeousEyes]

Nov08 no.6ii

 

[Asheeta]

Nov08 no.6ii

Check if it is the same qu http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_4.pdf

so the solution:
Using distance formula find V:
(1/2)(600+1000)*V = 20 000
therefore, V=25 m/s
I'm sure you'll be able to do the next part after this.