IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: Phosu on May 03, 2010, 08:21:39 am
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1)the points A, O and B lie on a straight horizontal track. A is 20m from O and B is on the other side of O at a distance of xm from O.
A-------------------O-----------------------B
20m xm
at time t=0 a particle P atarts from rest at O and moves towards B with uniform acceleration of 3m/s2. At the same instant, another particle Q, which is at the point A, is moving with a velocity 3m/s in the direction of O with uniform acceleration of 4m/s2 in the same direction.
Given that Q collides with P at B, find the value of x.
2)whilst looking over the edge of a vertical cliff, 122.5 metres in height, Jim dislodges a stone.
the stone falls freely from rest towards the sea below.
Ignoring the effect of air resistance,
a) calculate the time it would take for the stone to reache the sea
b) find the spreed with which the stone would hit the water.
Tow seconds after the stone begins to fall, Jim throws a tennis ball downwards at the stone. the tennis ball's initial speed is um/s. and it hits the stone before they both reach the water.
c) find the minimum value of u.
d) if u had taken air resistance into account in your calculations, what effect would this have had on your answer to part c) explain your ans.
plz solve and explain part c and d.
3 see img attached for diagram
Fig shows 2 particles P and Q, of mass 3kg and 2kg respectively, attached to the ends of a light inextensible string which passes over a smooth, fixd pulley. The system is released from rest with Pand Q at the same level 1.5m above the ground and 2 meters below the pulley
a) show that the inital acceleration of the system is g/5 m/s2
b) find the tension in the string
c) find the speed with which p hits the ground.
when p hits the ground it does not rebound
d) what is the closest that q gets to the pulley?
Plz explain how to do part D)
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Can you please tell me which past paper you got the question from ???
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umm i dont really know since we got these questions in a class test.
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could someone plzz do it...this is kinda urgent!
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sry for being late could u just provide us with the answers if u have them
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Give me 10min to ans dem.
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These r answered 2 da best of my knowledge,so get dem re-checked specially Q.2 part c. & d.
Q1> solve wit eqn s=ut+0.5at2
as wen dey meet, v shud find da time for which dey cover da same distance
3t+0.5*4*t2-20=0t+0.5*3*t2
0.5t2+32-20=0
t=4(+ve value only)
s(x)=0.5*3*42
s(x)=24m
Q2>a.
s=122.5 u=0 a=9.8 t=?
s=ut+0.5at2
122.5=0t+0.5*9.8*t2
t=5 sec
b.
s=122.5 u=0 a=9.8 v=?
v2=u2+2*a*s
v2=0+2*9.8*122.5
v=49m/s
c.(not sure)
min value of "u" wen objects meet at surface.
dre4 s=122.5 t=5+2(assuming u wrote two) a=9.8 u=?
s=ut+0.5at2
122.5=7u+0.5*9.8*72
u=12.6m/s
d.(not sure)
the Min value will depend of other factors, as time taken 4 da object 2 reach da surface will decrease.
Q3>a. & b.
3g-T=3a
T-2g=2a
if u substitute dem u get a=g/5 m/s(1.96) & T=23.52
c.
a=1.96m/s2 u=0 s=1.5 v=?
v2=u2+2as
v2=0+2*1.96*1.5
v2=5.88
d.
dats da max distance moved by Q wit its existing force.
u=(speed with which p hits the ground) a=-9.8 v=0 s=?
v2=u2+2*a*s
0=5.88+2*-9.8*s
s=0.3m(moved by q)
There4
2-0.3=1.7m closer 2 pully.
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srry for the typo, it is 2 sec
but why did u add the secs...?
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sry 4 da mistake,
It shud b t=5-2
=3
because as da stone reached da water surface at 5sec, da ball shut reach it at 3 sec, as its 2sec late.
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could someone plz explain this???
a car of mass 1000kg exerts a driving force of 2.2kN when pulling a caravan of mass 500kg along a horizontal road.
The car and caravan increase speed from rest to 4m/s while traveling 16m.
given that the resistances on the car and caravan are proptional to their masses, find these resistances and the tension in the tow bar.
could someone specifically explain what they mean by the red colour sentence.
thx ASAP plzzzz
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more the mass, more the resistance....
so the resistance of car will be more than that of the caravan.
well you could find the ratio of mass car:caravan, which is 2:1
find the total resistance force, and then divide the resistance in the ratio.
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give me 5min
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https://studentforums.biz/index.php/topic,5617.0.html
This is the exact same question asked before. Hope you find it useful.
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u=0 v=4 s=16 a=?
v2=u2+2as
42=2*a*16
a=0.8
F-R=ma
2200-R=(1000+500)*0.8
R=1000N
Ratio of mass 2:1
Car resistance=666.67 caravan resistance==333.33
T-R=ma
T-333.33=500*0.8
T=733.33
plz post da ms ans to b sure.
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thx all
but immortal only 733.3 is correct the rest is wrong :S
the answers:966.66 , and , 483.33 , and , 733.33
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thx all
but immortal only 733.3 is correct the rest is wrong :S
the answers:966.66 , and , 483.33 , and , 733.33
can u explain how i got da first 2 parts wrong, & got da last one correct. because if da first is wrong all shud b wrong.
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can u explain how i got da first 2 parts wrong, & got da last one correct. because if da first is wrong all shud b wrong.
i rili have no idea :s this is a question from my textbook, and they give no explanation
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the acceleration u calculated is rong
42=2xax16
16/32=a
a=0.5
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the acceleration u calculated is rong
42=2xax16
16/32=a
a=0.5
OMG ure write.
cant believe, i'm making dumb mistakes,wit exams coming up in a week.
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a particle A of mass 2.5kg which is at rest on a smooth inclined plane of angle 25 is connected to a particle B of mass 1.5kg by a light inextensible string which lies along a line of greatest slope of the plane and passes over a fixed smooth pulley at the top of the plane. B hangs freely and the system is released from rest with the string taut. find the acceleration of the system and the tension in the string. if the angle of inclination of the plane is increased to 45 find the new acceleration.
just how to do the part in red ??
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The inclination as resulted particle a moving down da hill, that's da only difference.so da ans cn b found similar 2 da first half which u would get a negative acceleration(deceleration)
T-1.5g=1.5a
2.5gsin25-T=2.5a
Substitute this eqn & u'll get ur ans.
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The inclination as resulted particle a moving down da hill, that's da only difference.so da ans cn b found similar 2 da first half which u would get a negative acceleration(deceleration)
T-1.5g=1.5a
2.5gsin25-T=2.5a
Substitute this eqn & u'll get ur ans.
thats exactly wat i did! just substituted the 45 angle instead of the previous angle in the above equations...but its different from the bk ! :S
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thats exactly wat i did! just substituted the 45 angle instead of the previous angle in the above equations...but its different from the bk ! :S
dude there should be downward a of 2.62 is that rite...........