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Mechanics 1 [Doubts and Solutions]

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This thread is about posting any doubt you have while doing mechanics 1. Don't be afraid to post any question, it doesn't matter how easy and how silly you might sound by posting such a question. Whats important is clearing your doubts. A collection of minds and a collection of different methodological solutions is what I aim to get from this thread. I hope that everyone would be helping each other as its one of the main purpose of this thread. So as to start and give a simple idea about how to go on about this thread, I will start off with a question that I came across.

Capture.JPG Those who know or those who think can solve this question go on give your best try and do the question. I won't reveal the answers until someone answers this correctly, and please be kind enough to show your working, [ imagine you are doing an Exam! ] Haha! Thanks in Advance :)!
 
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I HAVE A QUESTION WHICH I NEED AN ANSWER FOR: A BLOCK OF MASS 20KG IS RELEASED FROM REST AT THE TOP OF A ROUGH SLOPE. THE SLOPE IS INCLINED TO THE HORIZONTAL AT ANGLE OF 30. AFTER 6s THE SPEED OF THE BLOCK IS 21ms^-1. AS THE BLOCK SLIDES DOWN THE SLOPE IT IS SUBJECT TO A CONSTANT RESISTANCE OF MAGNITUDE RN. FIND THE VALUE OF R???? PLEASE I NEED THE ANSWER AS SOON AS YOU POSSIBLE?
 
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I HAVE A QUESTION WHICH I NEED AN ANSWER FOR: A BLOCK OF MASS 20KG IS RELEASED FROM REST AT THE TOP OF A ROUGH SLOPE. THE SLOPE IS INCLINED TO THE HORIZONTAL AT ANGLE OF 30. AFTER 6s THE SPEED OF THE BLOCK IS 21ms^-1. AS THE BLOCK SLIDES DOWN THE SLOPE IT IS SUBJECT TO A CONSTANT RESISTANCE OF MAGNITUDE RN. FIND THE VALUE OF R???? PLEASE I NEED THE ANSWER AS SOON AS YOU POSSIBLE?
wasnt there a diagram of the arrangements of force? it would have been easier then
 
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Two particles p and q of equal mass are connected by a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of a smooth inclined plane. The plane is inclined toy the horizontal at an angle 'a' where tan'a' = 0.75. Particle P is held at rest on the inclined plane at a distance of 2m from the pulley and Q hangs freely on the edge of the plane at a distance of 3m above the ground with the string vertical and taut.
Particle P is released. Find the speed with which it hits the pulley.
 
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JAN 2008, Question 6 part d.
At time t=0, P is at the point A with position vector (7i-1oj)m realtive to a fixed origin O. When t=3 s, where u and v are constant. after a further 4s, it passes through O and continues to move with velocity (ui + vj) ms-1

c) find the values of u and v

d) find the total time taken for P to move from A to a position which is due south of A.
 
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JAN 2008, Question 6 part d.
At time t=0, P is at the point A with position vector (7i-1oj)m realtive to a fixed origin O. When t=3 s, where u and v are constant. after a further 4s, it passes through O and continues to move with velocity (ui + vj) ms-1

c) find the values of u and v

d) find the total time taken for P to move from A to a position which is due south of A.


http://www.examsolutions.net/a-leve...worked-solution.php?paper_id=503&solution=6.4

This video will help you understand better :)! If not you can ask me further questions here.
 
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Ali Ahsan, I also need help in M1, I hope you would be kind enough to reply to my queries and if not then its fine, no biggies
Sure! :)! I will be glad to help! though I m quite busy, I would help you as soon as I get the time to check! Just ask me anything you have difficulty in.
 
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how to calculate the components of the resulatant of coplanar forces in horizontal and vertical directions. plz help
 
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how to calculate the components of the resulatant of coplanar forces in horizontal and vertical directions. plz help
Well the first thing you have to do is just simply resolve all the coplanar forces in the x-direction and y-direction, remember you are dealing with vectors, so keep in mind that direction matters, so taking into account the direction, there will be probably some negative and positive values, add all of them together.

All of this should be done separately as in, you have to resolve ALL the coplanar forces in the x-direction and then add all of them together to get the resultant x-direction component of the resultant force vector, then resolve ALL y-direction coplanar forces add them together to get the y-direction resultant component, now you have 'x' and 'y' direction components of the resultant force vector,
all you have to do now is simply use Pythagoras theorem to get the resultant force vector.
 
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Question 1 in Jan 2014.

A truck P of mass 2M is moving with speed U on smooth straight horizontal rails. It
collides directly with another truck Q of mass 3M which is moving with speed 4U in the
opposite direction on the same rails. The trucks join so that immediately after the collision
they move together. By modelling the trucks as particles, find
(a) the speed of the trucks immediately after the collision,
(b) the magnitude of the impulse exerted on P by Q in the collision.

The thing is that i have the right answers. Only my signs are wrong.
this is what i did

(a) 2MU - 12MU = v ( m1 + m2 )
===> v = -2U

but the ms says v = 2U only, and -2U is incorrect. and since i had the 'wrong' answer for (a), i also got the sign in (b) wrong. What is wrong with this?

Thank you! :D
 
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8. Two trains, A and B, start together from rest, at time t = 0, at a station and move along parallel straight horizontal tracks. Both trains come to rest at the next station after 180 s.
Train A moves with constant acceleration 2/3 m s–2 for 30 s, then moves at constant speed for 120 s and then moves with constant deceleration for the final 30 s.

Train B moves with constant acceleration for 90 s and then moves with constant deceleration for the final 90 s.

(a) Sketch, on the same axes, the speed–time graphs for the motion of the two trains
between the two stations.

(b) Find the acceleration of train B for the first half of its journey.

(c) Find the times when the two trains are moving at the same speed.

(d) Find the distance between the trains 96 s after they start.

can somebody PLEASE tell me how part (b) and (d) is done???
 
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8. Two trains, A and B, start together from rest, at time t = 0, at a station and move along parallel straight horizontal tracks. Both trains come to rest at the next station after 180 s.
Train A moves with constant acceleration 2/3 m s–2 for 30 s, then moves at constant speed for 120 s and then moves with constant deceleration for the final 30 s.

Train B moves with constant acceleration for 90 s and then moves with constant deceleration for the final 90 s.

(a) Sketch, on the same axes, the speed–time graphs for the motion of the two trains
between the two stations.

(b) Find the acceleration of train B for the first half of its journey.

(c) Find the times when the two trains are moving at the same speed.

(d) Find the distance between the trains 96 s after they start.

can somebody PLEASE tell me how part (b) and (d) is done???



so for part (b) u can equate the distance the two trains travel since they are the same. From the graph, one can see that the distance for train A is represented in a trapezium. Distance travelled= 0.5(h)(a+b)= 0.5*20*(180+120)=3000m.

Train B has a triangular area under its graph.
0.5*base*height= 0.5*180*v=3000
v=33.3 m/s

It accelerates for 90s and thats the slope of the first 90s graph.
a=(33.3-0)/90-0)
a=0.37 ms^-2
 
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8. Two trains, A and B, start together from rest, at time t = 0, at a station and move along parallel straight horizontal tracks. Both trains come to rest at the next station after 180 s.
Train A moves with constant acceleration 2/3 m s–2 for 30 s, then moves at constant speed for 120 s and then moves with constant deceleration for the final 30 s.

Train B moves with constant acceleration for 90 s and then moves with constant deceleration for the final 90 s.

(a) Sketch, on the same axes, the speed–time graphs for the motion of the two trains
between the two stations.

(b) Find the acceleration of train B for the first half of its journey.

(c) Find the times when the two trains are moving at the same speed.

(d) Find the distance between the trains 96 s after they start.

can somebody PLEASE tell me how part (b) and (d) is done???


For (d) calculate distance travelled by each of the trains and take the difference.

Train A= 0.5bh (for the first 30s) + wb (for the next 66s)
(0.5*30*20) + (66*20)
=1620 m

Train B= 0.5bh ( for the first 90s) + 0.5(h)(a+b) (for the next 6s)
(0.5*90*33.3) + 0.5*6*(33.3+31.08)
=1691.64 m

Seems like Train B is faster hey. So the difference is 71.64 m :) :)
 
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