M1 doubts

[Phosu]

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)

[Saladin]

Can you please tell me which past paper you got the question from

[Phosu]

umm i dont really know since we got these questions in a class test.

[Phosu]

could someone plzz do it...this is kinda urgent!

[7ooD]

sry for being late could u just provide us with the answers if u have them

[immortal]

Give me 10min to ans dem.

[immortal]

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.

[Phosu]

srry for the typo, it is 2 sec
but why did u add the secs...?

[immortal]

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.

[halosh92]

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

[Phosu]

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.

[immortal]

give me 5min

[Saladin]

https://studentforums.biz/index.php/topic,5617.0.html

This is the exact same question asked before. Hope you find it useful.

[immortal]

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.

[halosh92]

thx all
but immortal only 733.3 is correct the rest is wrong :S
the answers:966.66        , and     ,       483.33 ,             and ,      733.33