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help with math mechanics paper4 m/j 02 question!!!!!!please!
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Re: help with math mechanics paper4 m/j 02 question!!!!!!ple
In (ii), use the 1st equation of motion v(f) = v(i) + at
= 0 + 2.5(2)
= 5m/s (the velocity of A upwards b4 string becomes slack)
Then, again use the 1st equation to find time that A takes to move upwards b4 stopping momentarily, using
v(f) = v(i) + gt
0 = 5 + (-10)(t)
----------> t = 0.5s
Now use the idea that time for upward motion = time for downward motion, for A in this case, assuming negligible air resistance
i.e. total time for the motion = 2 x time for upward motion (for A)
= 2 x 0.5
= 1s
When A returns to its starting position* (where the string became slack), the string will become taut again. Therefore, the required time = 1s
For part (ii), 1st draw a t(time)-axis as x-axis while a v(velocity)-axis as y-axis.
Then decide which direction (upwards or downwards) u want to take as positive and which one as negative. Here (btw the convention is to take upward as +ve) i am taking upwards as positive.
Therefore, the graph for B will be a straight line with a -ve gradient of magnitude 2.5m/s^2 for 2 seconds. The following point is not included in MS bt i think it shud b @ The next moment, it will suffer an infinite deceleration, making the velocity 0m/s (since it hits the floor and comes to rest). Now, since B doesn't rebound, its velocity will remain zero.
Whereas, for A it will be a straight line with a +ve gradient of magnitude 2.5m/s^2 for the 1st 2 seconds. But the next moment (after 2 seconds) it will have a -ve gradient of g(~10)m/s^2, taking the velocity of A down to 0m/s (when it stops momentarily) in 0.5s, and then increases -vely (as it's now heading downwards to its starting position*) for another 0.5s. This completes the graph
I hope this helps
In (ii), use the 1st equation of motion v(f) = v(i) + at
= 0 + 2.5(2)
= 5m/s (the velocity of A upwards b4 string becomes slack)
Then, again use the 1st equation to find time that A takes to move upwards b4 stopping momentarily, using
v(f) = v(i) + gt
0 = 5 + (-10)(t)
----------> t = 0.5s
Now use the idea that time for upward motion = time for downward motion, for A in this case, assuming negligible air resistance
i.e. total time for the motion = 2 x time for upward motion (for A)
= 2 x 0.5
= 1s
When A returns to its starting position* (where the string became slack), the string will become taut again. Therefore, the required time = 1s
For part (ii), 1st draw a t(time)-axis as x-axis while a v(velocity)-axis as y-axis.
Then decide which direction (upwards or downwards) u want to take as positive and which one as negative. Here (btw the convention is to take upward as +ve) i am taking upwards as positive.
Therefore, the graph for B will be a straight line with a -ve gradient of magnitude 2.5m/s^2 for 2 seconds. The following point is not included in MS bt i think it shud b @ The next moment, it will suffer an infinite deceleration, making the velocity 0m/s (since it hits the floor and comes to rest). Now, since B doesn't rebound, its velocity will remain zero.
Whereas, for A it will be a straight line with a +ve gradient of magnitude 2.5m/s^2 for the 1st 2 seconds. But the next moment (after 2 seconds) it will have a -ve gradient of g(~10)m/s^2, taking the velocity of A down to 0m/s (when it stops momentarily) in 0.5s, and then increases -vely (as it's now heading downwards to its starting position*) for another 0.5s. This completes the graph
I hope this helps
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Re: help with math mechanics paper4 m/j 02 question!!!!!!ple
p4 o/n 01 question 6 (ii) & (iii)
for part (ii)
When Q hits the ground, P is ready to start free-fall (in upward direction) with an initial velocity, say u. We can figure this u out using the fact that Q and P travel the same distance of 4.5m when the string is taut.
2as = v(f)^2 - v(i)^2
Now the v(i) is 0 since P was initially at rest (system was at rest); a = [g(=10)/9] m/s^2; h = 4.5m -----------------------------> v(f) = sqrt(10) m/s = u.
Now, assuming upward direction is +ve while downwards -ve
v(f) = v(i) + at
0 = sqrt(10) + [-g(=-10m/s^2)] * t(?)
----> t(?) = 1/sqrt(10) s ~ 0.316s
for part (iii),
Use the fact that P travels the same distance upwards AND downwards, its initial velocity is sqrt(10) m/s and its flight time (to travel to max height) is 1/sqrt(10)
s(P) = distance travelled upwards + distance travelled upwards
= 2 x distance travelled upwards
Using the equation: s = 0.5[v(i) + v(f)]t
s(P) = 2 x 0.5[sqrt(10) + 0][1/sqrt(10)]
= 1m
p4 o/n 01 question 6 (ii) & (iii)
for part (ii)
When Q hits the ground, P is ready to start free-fall (in upward direction) with an initial velocity, say u. We can figure this u out using the fact that Q and P travel the same distance of 4.5m when the string is taut.
2as = v(f)^2 - v(i)^2
Now the v(i) is 0 since P was initially at rest (system was at rest); a = [g(=10)/9] m/s^2; h = 4.5m -----------------------------> v(f) = sqrt(10) m/s = u.
Now, assuming upward direction is +ve while downwards -ve
v(f) = v(i) + at
0 = sqrt(10) + [-g(=-10m/s^2)] * t(?)
----> t(?) = 1/sqrt(10) s ~ 0.316s
for part (iii),
Use the fact that P travels the same distance upwards AND downwards, its initial velocity is sqrt(10) m/s and its flight time (to travel to max height) is 1/sqrt(10)
s(P) = distance travelled upwards + distance travelled upwards
= 2 x distance travelled upwards
Using the equation: s = 0.5[v(i) + v(f)]t
s(P) = 2 x 0.5[sqrt(10) + 0][1/sqrt(10)]
= 1m