• 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)

Mathematics: Post your doubts here!

S

Sweet Princess

Messages
8
Reaction score
0
Points
1
Can anybody pls solve these for me...: 9709/04/M/J/04 Q.4(ii) and Q.7(iii)......
please help.....i also asked before but nobody replied......
 
H

hm12

Messages
182
Reaction score
29
Points
38
Sweet Princess said:
Can anybody pls solve these for me...: 9709/04/M/J/04 Q.4(ii) and Q.7(iii)......
please help.....i also asked before but nobody replied......
Click to expand...
question 4 part (ii) find the angle of inclination by sin angle=0.7/2.5 so angle =16.3 now if you resolve the forces on the mass then normal contact force would be equal to 2cos16.3=1.92 now frictional force=1.92*0.15=0.288 workdone by frictional forces= 0.288*2.5=0.72 now KE=PE-WD=1.4-0.72=0.68J
 
H

hm12

Messages
182
Reaction score
29
Points
38
Sweet Princess said:
Can anybody pls solve these for me...: 9709/04/M/J/04 Q.4(ii) and Q.7(iii)......
please help.....i also asked before but nobody replied......
Click to expand...
question 7 part (iii) at t=1.25s P2 and P1 are at the same height but P2 is moving downwards and P1 upwards so if you minus 1.25 from the total time P1 takes to reach its maximum height you will get the answer v=u+at so 0=30-10t ---->t=3s (at max height v=0) so time it is above P2 =3-1.25=1.75s
 
S

Sweet Princess

Messages
8
Reaction score
0
Points
1
hm12 said:
question 7 part (iii) at t=1.25s P2 and P1 are at the same height but P2 is moving downwards and P1 upwards so if you minus 1.25 from the total time P1 takes to reach its maximum height you will get the answer v=u+at so 0=30-10t ---->t=3s (at max height v=0) so time it is above P2 =3-1.25=1.75s
Click to expand...
Thankuuu.... :)
 
H

hm12

Messages
182
Reaction score
29
Points
38
Sweet Princess said:
but the formula of WD is F*d*(cosα)....so why we haven't taken dcosα ??
Click to expand...
what? no where did the cosα came from? i don't get that? as far as i know there is no cosα in WD formula sorry i cant explain that much i am not too good at mechanics
 
Soldier313

Soldier313

Messages
960
Reaction score
3,500
Points
253
hm12 said:
please explain the whole thing :)
Click to expand...
okay so
part i.) u use (v)*2=(u)*2 +2as substitute the values given in the qn and u get 7m/s as ur ans.
part ii.) u use this mgsin(theta) x distance - (work done against friction) = ma x distance
then u substitute the values provided and make theta the subject and you will get the angle as 21.1
part iii.) average speed = (u + v)/2 =(3+7)/2 = 5m/s then u use this in the equation (v)*2 = (u)*2 +2as and substitute 5 in place of v and the rest of the values substituted a before....u then use this:
the ratio of work done = the ratio of distances...
thus using previous information getting : WD against friction/7 = 3.2/8
hope you understood:)
 
You must log in or register to reply here.
Top