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    Here is it, need help in only part B
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    (Original post by lil_jack)
    Here is it, need help in only part B
    What have you tried? Please post your working / thoughts.
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    (Original post by notnek)
    What have you tried? Please post your working / thoughts.
    I tried part A and got acceleration to be 0.4m/s^2 but i have no idea how to attempt part B
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    (Original post by lil_jack)
    I tried part A and got acceleration to be 0.4m/s^2 but i have no idea how to attempt part B
    First you need to understand what will happen to A and why it will stop:

    A will start from rest and have the acceleration from part a) so will move up the slope. Once B hits the ground (A will have traveled 2m), the string will become slack and the only acceleration will be due to the weight component of A down the slope. So A will decelerate and eventually come to rest.

    Does this all make sense?


    So you can divide the motion of A up into two : the initial motion when the string is taut and the secondary motion when the string becomes slack.

    For the initial motion you have

    u = 0
    a = (found in part a))
    s = 2

    Start by finding the final speed (v) of this 2m movement.


    Try this first and post your working if you get stuck or don't know what to do next.
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    I found v=(2√10)/5, not sure what i can do next
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    (Original post by lil_jack)
    I found v=(2√10)/5, not sure what i can do next
    I haven't checked it but lets assume it's correct for now.

    Do you understand the motion of the particle? There's no point doing anymore work if you don't.

    So now you have the final speed of the initial part of the motion which is the same as the initial speed for the second part of the motion.

    The next step is to work out the acceleration of A once B hits the floor and the string becomes slack. At this point the only force acting on A will be the component of A's weight acting down the slope (there's no friction). Can you use this to work out the acceleration?

    Then you can use this acceleration with v = 0 to find the distance moved by A before coming to rest.
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    I think I have misunderstanding in the motion of this particle, that why i cant do anything further
    as i what i thought simply A will have moved 2m (which is the distance that the particle B moved to reach ground) and comes to rest. Thanks alot anyway for your effort and I will try to visualize this hopefully will get the answer
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    (Original post by lil_jack)
    I think I have misunderstanding in the motion of this particle, that why i cant do anything further
    as i what i thought simply A will have moved 2m (which is the distance that the particle B moved to reach ground) and comes to rest. Thanks alot anyway for your effort and I will try to visualize this hopefully will get the answer
    I understand your confusion. You need to try to imagine this happening in real life. 'A' was already moving up the slope so if 'B' hit the floor, it would be weird if 'A' suddenly stopped. 'A' had some speed before 'B' hit the floor so will continue to move up the slope.
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    well after B have reached ground and only force acting now was 15Sin(36.86), I used f=ma ,,0-15sin(36.86)=1.5a and found acceleration to be -6m/s^2 and By having the following
    a=-6
    u=(2√10)/5
    v=0
    and s=?
    0=((2√10)/5)^2+2*-6*S
    S=0.133
    so total distance moved tell it came to rest is = 2+0.13=2.13
    Is that roughly right by looking at it?
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    (Original post by lil_jack)
    well after B have reached ground and only force acting now was 15Sin(36.86), I used f=ma ,,0-15sin(36.86)=1.5a and found acceleration to be -6m/s^2 and By having the following
    a=-6
    u=(2√10)/5
    v=0
    and s=?
    0=((2√10)/5)^2+2*-6*S
    S=0.133
    so total distance moved tell it came to rest is = 2+0.13=2.13
    Is that roughly right by looking at it?
    At first glance your working looks good but I haven't checked any of your numbers. Do you have the answer?

    I can give it a go later if you want to know if your answer is correct.

    By the way in M1 you'll meet these types of questions a lot in exams where there are two parts to the motion. If you properly understand this question then it's a good sign that you'll be able to do future questions like this.
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    yeahh i just found the answer in the pdf answers file to be 2.13m and it is exactly like mine
    Thanks alot for your explanation and for your time
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    Just a small tip to add about the angle QPR. In the diagram you have the opposite (3m) and the hypotenuse (5m), so you can write down Sin(PQR)=3/5 without working out what the angle is in degrees.

    Furthermore, as it's a right angled triangle you can calculate the length of side PR using Pythagoras. If you know the length of all three sides, you could write down the sine, cosine and tangent of QPR (if you needed them) as fractions without referring to the angle in degrees. M1 and M2 questions are often designed so that you can do this.
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    (Original post by old_engineer)
    Just a small tip to add about the angle QPR. In the diagram you have the opposite (3m) and the hypotenuse (5m), so you can write down Sin(PQR)=3/5 without working out what the angle is in degrees.

    Furthermore, as it's a right angled triangle you can calculate the length of side PR using Pythagoras. If you know the length of all three sides, you could write down the sine, cosine and tangent of QPR (if you needed them) as fractions without referring to the angle in degrees. M1 and M2 questions are often designed so that you can do this.
    Very good tip. I didn't see the lengths given in the question and assumed they'd been given the angle.
 
 
 
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