Qualification > Sciences
ALL CIE PHYSICS DOUBTS HERE !!!
WARRIOR:
--- Quote from: $!$RatJumper$!$ on October 30, 2010, 06:16:19 pm ---Could you tell which paper these questions are from? it would make it easier.
--- End quote ---
its questions from my school book :/
missbeautiful789:
For mass m1:
Resultant force = mass x acceleration
The only force on m1 is the tension on the string. Its weight is being "neutralised" by the reaction force of the table, so forget about its weight.
so
T=3xa
For mass M2:
Its hanging, so the string is pulling it upward and its weight is acting downward. So its RF= T-mg
assuming g=10
T-20=2xa
T=2xa+20
T is the same so we can say:
3a=2a+20
a=20
B) final velocity=initial velocity + acceleration x time
v=u+at
from rest, so u=o
V=at
v=20x1.2
V=24
c) S=ut+0.5xaxt2
s=0.5x20x1.22
s=14.4
d) s=0.5x20x0.42
s=1.6
______________________________________________
2500=1400xa
a=1.79
_________________________________________
resolved force of weight down plane= mg sin@
v2=u2+2aS
625=2xax50
a=6.25
RF=ma so
mg sin@=mx6.25
mx10xsin@=mx6.25
sin@=0.625
@=38.7 degrees
Looks like homework..don't forget the units
WARRIOR:
--- Quote from: missbeautiful789 on October 30, 2010, 09:19:21 pm ---For mass m1:
Resultant force = mass x acceleration
The only force on m1 is the tension on the string. Its weight is being "neutralised" by the reaction force of the table, so forget about its weight.
so
T=3xa
For mass M2:
Its hanging, so the string is pulling it upward and its weight is acting downward. So its RF= T-mg
assuming g=10
T-20=2xa
T=2xa+20
T is the same so we can say:
3a=2a+20
a=20
B) final velocity=initial velocity + acceleration x time
v=u+at
from rest, so u=o
V=at
v=20x1.2
V=24
c) S=ut+0.5xaxt2
s=0.5x20x1.22
s=14.4
d) s=0.5x20x0.42
s=1.6
______________________________________________
2500=1400xa
a=1.79
_________________________________________
resolved force of weight down plane= mg sin@
v2=u2+2aS
625=2xax50
a=6.25
RF=ma so
mg sin@=mx6.25
mx10xsin@=mx6.25
sin@=0.625
@=38.7 degrees
Looks like homework..don't forget the units
--- End quote ---
yes it was infact hw!
thanks ! + rep
that last one was pretty hard for me! even with your answers inftont of me
thanks again
WARRIOR:
yo guys !
i have a tension test and a vector and projectiles test
if anyone has any exams or notes or question and can be kind enough to post them..CHEERS!
Deadly_king:
--- Quote from: Salahuddin_Ansari on October 30, 2010, 09:19:21 pm ---For mass m1:
Resultant force = mass x acceleration
The only force on m1 is the tension on the string. Its weight is being "neutralised" by the reaction force of the table, so forget about its weight.
so
T=3xa
For mass M2:
Its hanging, so the string is pulling it upward and its weight is acting downward. So its RF= T-mg
assuming g=10
T-20=2xa
T=2xa+20
T is the same so we can say:
3a=2a+20
a=20
B) final velocity=initial velocity + acceleration x time
v=u+at
from rest, so u=o
V=at
v=20x1.2
V=24
c) S=ut+0.5xaxt2
s=0.5x20x1.22
s=14.4
d) s=0.5x20x0.42
s=1.6
--- End quote ---
She made just a small mistake there.
Actually the tension is upwards towards the pulley, but the acceleration is downwards due to the weight of M2.
Hence the equation should have been 20 - T = 2a
Now you can solve it with the first equation T = 3a and you'll be getting the acceleration to be 4ms-2
b) All the other answers were affected since the acceleration was not correct. Otherwise the method was perfect ;)
v = u + at ----> v = 0 + 4(1.2) = 4.8ms-1
c) v2 = u2 + 2aS ----> 4.82= 0 + 2(4)S
Hence the distance S is found to be 2.88m
d) Displacement is given by S = ut + 0.5at2
S = 0 + 0.5(4)(0.4)2 ---> S = 0.32m
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