Even if passing through a pulley and it goes to some other angle?Yes, if and only if no external force is acting and if the string is vertical
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Even if passing through a pulley and it goes to some other angle?Yes, if and only if no external force is acting and if the string is vertical
Also. Why is C never taken when integrating velocity to get displacementsYes, if and only if no external force is acting and if the string is vertical
i need help iin this tooHey Guys, can someone please solve 6 (ii)?
Hey Guys, can someone please solve 6 (ii)?
Aren't we supposed to make the graph in the negative section? because it is going downwards isn't it?there are 2 cases of constant acceleration.
Before reaching the liquid:
a = 10 s = 5 u = o v = ? t = ?
use constant acceleration formulas to find v and t
v = 10, t = 1
Inside the liquid:
a = 5.5 s = 4 u = 10 (v of previous) v = ? t = ?
again use constant acceleration formulas
v = 12, t = 0.36
For the sketch, make a straight line from the origin to (1, 10). This is for the part before the liquid
Then make a line of less gradient from (1,10) to (1.36, 12) for the part inside the liquid.
It doesn't matter. You can take any direction as positive as long as the other direction is considered negative.Aren't we supposed to make the graph in the negative section? because it is going downwards isn't it?
can u plz draw the graphHey Guys, can someone please solve 6 (ii)?
Won't they cut marks if I drew it in the wrong direction?It doesn't matter. You can take any direction as positive as long as the other direction is considered negative.
No it will be downwards if it had bounced off (Part after bouncing will be taken as downward). In this case it is only moving in one directionAren't we supposed to make the graph in the negative section? because it is going downwards isn't it?
Do u know how to convert a velocity time graph to distance time or vice versa? I'm rly facing issue in thisIt doesn't matter. You can take any direction as positive as long as the other direction is considered negative.
No it will be downwards if it had bounced off (Part after bouncing will be taken as downward). In this case it is only moving in one direction
okay what if the ball was for example projected upwards how will the graph look like?It doesn't matter. You can take any direction as positive as long as the other direction is considered negative.
Yes, if the pulley is smooth , it is the same string so same tensionEven if passing through a pulley and it goes to some other angle?
which year is this ?Why isn't T2 20s? My values were 10 and 30 but when I inserted value of 21 in second derivative of velocity it was giving me positive so increasing? And when I put 19 it was giving me negative so decreasing?View attachment 53464
Thanks broYes, if the pulley is smooth , it is the same string so same tension
which year is this ?
can u draw the graphYes, if the pulley is smooth , it is the same string so same tension
Also upward.okay what if the ball was for example projected upwards how will the graph look like?
Why isn't T2 20s? My values were 10 and 30 but when I inserted value of 21 in second derivative of velocity it was giving me positive so increasing? And when I put 19 it was giving me negative so decreasing?View attachment 53464
It's not too bad as long as you understand that the v/t graph is the graph of the gradient of s/t and remember a few cases.Do u know how to convert a velocity time graph to distance time or vice versa? I'm rly facing issue in this
Because we are doing definite integration. The c cancels out.I have some doubts please clear.
- Why is 'c' ignored when integrating velocity to find disp?
There is no 'right' or 'wrong' direction. Positive can be any direction as long as you show motion in the opposite direction as negative.Won't they cut marks if I drew it in the wrong direction?
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