Mechanics question help

9

https://imgur.com/a/7qvHJXV

Could someone help me with part b please? from part a I got 0.4g as the acceleration and think I will use suvat but I don’t have s for either one. I think that maybe I need to use simultaneous equations as b will move the same distance as a so try to do suvat for both but I get stuck there. Also as it’s 3 marks not sure that is the method to use

Reply 1

mqb2766

21

Original post

by username79352

https://imgur.com/a/7qvHJXV
Could someone help me with part b please? from part a I got 0.4g as the acceleration and think I will use suvat but I don’t have s for either one. I think that maybe I need to use simultaneous equations as b will move the same distance as a so try to do suvat for both but I get stuck there. Also as it’s 3 marks not sure that is the method to use

The string is 1m long and B starts 0.5m below the pulley so ...

You should check whether A hits the pulley before/after B hits the ground as the string will go slack (zero acceleration for A) after B hits the ground.

Note it usually helps to see any working/attempt at this/previous question parts.

Reply 2

username79352

OP

9

Original post

by mqb2766

The string is 1m long and B starts 0.5m below the pulley so ...
You should check whether A hits the pulley before/after B hits the ground as the string will go slack (zero acceleration for A) after B hits the ground.
Note it usually helps to see any working/attempt at this/previous question parts.

I completely missed the fact that b was 0.5m long, thank you for pointing that out. I’ll give it a try now.

Reply 3

username79352

OP

9

https://imgur.com/a/75htvGC

could you help me with part d please, my working out is attached above. I know the time that a hits the pulley and when I try to find the time taken for b to hit the ground by doing the time for b to drop once the string breaks, I get confused there and end up with a negative answer.

Reply 4

mqb2766

21

Youve overcomplicated d). The first part is unnecessary, you simply want another suvat phase where a=g,s=0.3,u_new=v_old as you say. You seem to have it right but its a parabola/quadratic function of time so youd expect two solutions. One is negative, so play the parabola backwards in time but you want the positive one. Note the c/a constant is negative so the two roots have opposite signs
https://www.desmos.com/calculator/zeho0nzyt9
The positive root is about 0.1s which seems about right.

Reply 5

username79352

OP

9

Original post

by mqb2766

Youve overcomplicated d). The first part is unnecessary, you simply want another suvat phase where a=g,s=0.3,u_new=v_old as you say. You seem to have it right but its a parabola/quadratic function of time so youd expect two solutions. One is negative, so play the parabola backwards in time but you want the positive one. Note the c/a constant is negative so the two roots have opposite signs
https://www.desmos.com/calculator/zeho0nzyt9
The positive root is about 0.1s which seems about right.

They say the answer is 0.250 not what I got. If I use s as 0.8 in my equation then I get that to 3sf but I don’t understand why I would use that value because when the string breaks the height of b is 0.3m from the ground.

Reply 6

mqb2766

21

Original post

by username79352

They say the answer is 0.250 not what I got. If I use s as 0.8 in my equation then I get that to 3sf but I don’t understand why I would use that value because when the string breaks the height of b is 0.3m from the ground.

Its 0.25s if you assume u=0 (and s=0.3), rather than the velocity that A hits the pulley with. The speed of A is obviously reduced to zero when it hits the pulley, though I cant see anything in the question to assume its the same for B, so the speed is reduced to zero, then the string breaks, then ....

Reply 7

username79352

OP

9

Original post

by mqb2766

Its 0.25s if you assume u=0 (and s=0.3), rather than the velocity that A hits the pulley with. The speed of A is obviously reduced to zero when it hits the pulley, though I cant see anything in the question to assume its the same for B, so the speed is reduced to zero, then the string breaks, then ....

Sorry I meant to say that in the textbook it says the answer is 0.250 to 3sf and when I use u as 0 and s as 0.3 I get 0.2474….

Reply 8

mqb2766

21

Original post

by username79352

Sorry I meant to say that in the textbook it says the answer is 0.250 to 3sf and when I use u as 0 and s as 0.3 I get 0.2474….

If thats the case neither ans makes a great deal of sense and the question is somewhat ill posed. It cant be 0.8m above the table when the string breaks, and I agree to 3sf its 0.247s when it falls 0.3m from a zero initial velocity. Really what the initial velocity of B would be after the break would depend on the string. If the string was "a thread", so just strong enough to not break during the normal motion of A sliding, then it would break "immediately" when A hit the pulley and the initial velocity of B would be largely unaffected. The question hints at this by asking you to calculate the velocity of A when it hits the pulley in the proceeding part. However, if the string was strong enough to impart an impulse on B large enough to reduce its velocity to zero and then break, the initial velocity of B would be zero for the freefall phase. I dont think there is a standard assumption about what happens for this scenario at a level, and Id have expected the question to give a hint and without any futher info, Id still have taken it to be A's velocity when it hits the pulley.