The Student Room Group

Mechanics inclination problem

Qu 1. Find the time interval between a particle reaching the bottom of a smooth slope of length 5m and inclination 1 in 98, and another particle reaching the bottom of s smooth slope of length 6m and incline 1 in 70. Both particles are released from rest at the top of their respective slopes at the same time.

Qu 2. A mass of 15kg lies on a smooth plane of inclination 1 in 49. One end of a light inextensible string is attached to this mass and the string passes up the line of greatest slope, over a smooth pulley fixed at the top of the plane and has freely suspended a mass of 10kg at its other end.
If the system is releases from rest, find the acceleration of the masses and the distance each travels in the first 2s. Assume that nothing impedes the motion of either mass.

Any attempts are welcome!!

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Original post by Gemma_98
Qu 1. Find the time interval between a particle reaching the bottom of a smooth slope of length 5m and inclination 1 in 98, and another particle reaching the bottom of s smooth slope of length 6m and incline 1 in 70. Both particles are released from rest at the top of their respective slopes at the same time.

Qu 2. A mass of 15kg lies on a smooth plane of inclination 1 in 49. One end of a light inextensible string is attached to this mass and the string passes up the line of greatest slope, over a smooth pulley fixed at the top of the plane and has freely suspended a mass of 10kg at its other end.
If the system is releases from rest, find the acceleration of the masses and the distance each travels in the first 2s. Assume that nothing impedes the motion of either mass.

Any attempts are welcome!!


q1 well start with a diagram with 98 as length on bottom and 1 as height and fill in what you know... wait i'm confused :/


i thought 1 in 98 meant that 1 is the height and 98 is the length of the flat bit of the slope touching the ground.... but how does that work if the slope length is 5m??? :/
(edited 7 years ago)
Reply 2
Original post by thefatone
q1 well start with a diagram with 98 as length on bottom and 1 as height and fill in what you know... wait i'm confused :/


i thought 1 in 98 meant that 1 is the height and 98 is the length of the flat bit of the slope touching the ground.... but how does that work if the slope length is 5m??? :/


It just means the angle is arctan(1/98)\arctan(1/98) - you can help out with the rest? I'm off to bed now.
Original post by Zacken
It just means the angle is arctan(1/98)\arctan(1/98) - you can help out with the rest? I'm off to bed now.


oh... i haven't done M2 or C3 or 4 yet :/
Reply 4
Original post by thefatone
q1 well start with a diagram with 98 as length on bottom and 1 as height and fill in what you know... wait i'm confused :/


i thought 1 in 98 meant that 1 is the height and 98 is the length of the flat bit of the slope touching the ground.... but how does that work if the slope length is 5m??? :/




I think the 1 in 98 is suppose to mean for every 1m it climbs its 98 along the bottom which allows you to figures out the angle to be
Sin theta = o/h = 1/98
once again my good intentions are useless ;( ;(
Reply 6
Original post by Gemma_98
I think the 1 in 98 is suppose to mean for every 1m it climbs its 98 along the bottom which allows you to figures out the angle to be
Sin theta = o/h = 1/98


The answer is a = 3.8ms2 and distance = 7.6m
Reply 7
Original post by thefatone
oh... i haven't done M2 or C3 or 4 yet :/


Somebody else will take care of it or I'll do it in a bit if nobody has yet. OP may wish to express his/her own thoughts in the mean time.
Reply 8
Original post by Gemma_98
I think the 1 in 98 is suppose to mean for every 1m it climbs its 98 along the bottom which allows you to figures out the angle to be
Sin theta = o/h = 1/98


Nah, tan theta = opp/adj = 1/98.

To start off: resolve down the plane, find accleeration (mg sin theta) then use SUVAT.
Original post by Zacken
Somebody else will take care of it or I'll do it in a bit if nobody has yet. OP may wish to express his/her own thoughts in the mean time.


i wish i knew more... ok then :smile:
Reply 10
The answer to qu1 is 0.74 secs but i dunno how this is
Reply 11
Original post by Gemma_98
Qu 1. Find the time interval between a particle reaching the bottom of a smooth slope of length 5m and inclination 1 in 98, and another particle reaching the bottom of s smooth slope of length 6m and incline 1 in 70. Both particles are released from rest at the top of their respective slopes at the same time.

Qu 2. A mass of 15kg lies on a smooth plane of inclination 1 in 49. One end of a light inextensible string is attached to this mass and the string passes up the line of greatest slope, over a smooth pulley fixed at the top of the plane and has freely suspended a mass of 10kg at its other end.
If the system is releases from rest, find the acceleration of the masses and the distance each travels in the first 2s. Assume that nothing impedes the motion of either mass.


Any attempts are welcome!!


Original post by Zacken
Nah, tan theta = opp/adj = 1/98.

To start off: resolve down the plane, find accleeration (mg sin theta) then use SUVAT.


Exactly what Zacken said,
Here's what I got for q1, if you don't understand any of the steps feel free to ask :h:

Spoiler

(edited 7 years ago)
Original post by KaylaB
Exactly what Zacken said,
Here's what I got for q1, if you don't understand any of the steps feel free to ask :h:

Spoiler



Give me your handwriting please
Reply 13
Original post by Student403
Give me your handwriting please


You can keep it! It's a mess! :lol::lol:
Original post by KaylaB
You can keep it! It's a mess! :lol::lol:


You only realise what you had once it's gone :3
Reply 15
Original post by Gemma_98
The answer to qu1 is 0.74 secs but i dunno how this is


This one I'm not as confident with the answer, but here's my attempt anywho

Spoiler

Reply 16
Original post by KaylaB
This one I'm not as confident with the answer, but here's my attempt anywho


You've forgotten the tension in the string. Sorry!
Reply 17
Original post by Zacken
You've forgotten the tension in the string. Sorry!


I still seem to to be getting the same value for a, and thus the final answer :s-smilie:
Reply 18
Original post by KaylaB
Exactly what Zacken said,
Here's what I got for q1, if you don't understand any of the steps feel free to ask :h:

Spoiler




Thank you!
I get the resolving and SUVAT but how did tan 1/98 become the sqr root thing?
Reply 19
Original post by Gemma_98
Thank you!
I get the resolving and SUVAT but how did tan 1/98 become the sqr root thing?


Well if we have the Opposite and the Adjacent of a right-angled triangle we can use Pythagoras to find the length of the Hypotenuse

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