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Author Topic: M1 doubts  (Read 3418 times)

Offline immortal

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Re: M1 doubts
« Reply #15 on: May 04, 2010, 01:10:41 pm »
Quote from: halosh92 on May 04, 2010, 12:54:10 pm
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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Offline halosh92

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Re: M1 doubts
« Reply #16 on: May 04, 2010, 01:16:16 pm »
Quote from: immortal on May 04, 2010, 01:10:41 pm
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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Offline Phosu

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Re: M1 doubts
« Reply #17 on: May 04, 2010, 01:58:36 pm »
the acceleration u calculated is rong

42=2xax16
16/32=a
a=0.5
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Offline immortal

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Re: M1 doubts
« Reply #18 on: May 04, 2010, 06:59:01 pm »
Quote from: Phosu on May 04, 2010, 01:58:36 pm
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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Offline halosh92

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Re: M1 doubts
« Reply #19 on: May 04, 2010, 07:11:41 pm »
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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Offline immortal

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Re: M1 doubts
« Reply #20 on: May 04, 2010, 10:48:23 pm »
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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Offline halosh92

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Re: M1 doubts
« Reply #21 on: May 05, 2010, 06:27:39 pm »
Quote from: immortal on May 04, 2010, 10:48:23 pm
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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Offline cashem'up

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Re: M1 doubts
« Reply #22 on: May 06, 2010, 10:54:03 am »
Quote from: halosh92 on May 05, 2010, 06:27:39 pm
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...........
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