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1. Any help would be great thanks.

Calculate the magnitude of the impulse of a force magnitude FN acting from thime t 1 seconds to time t2 seconds where.

F = 5 + e^-2t t1 =0 t2= 3

and

F=3cos5t t1 = 0 t2 = Pi/10
2. (Original post by droid)
Any help would be great thanks.

Calculate the magnitude of the impulse of a force magnitude FN acting from thime t 1 seconds to time t2 seconds where.

F = 5 + e^-2t t1 =0 t2= 3

and

F=3cos5t t1 = 0 t2 = Pi/10
In each case, integrate the function and put in t2 as top limit, t1 as bottom limit.

Aitch
3. I also have a question.
A particle of mass 2kg is hanging in equilibrium attached to one end of a light elastic string of natural length 1.2m and modulus 48N. P is pulled downwards a distance 0.5m and released. When P has risen a distance 0.3m, the string is cut. Calculate the greatest height P reaches above the equilibrium point and time taken for it to reach that given height.

I get the height correct (0.0143m), but the time i get as 0.411s :'( and not 0.468s as the book says
4. When the string is cut, P is a projectile. So consider vertical motion:
v=0, u=sqrt(4.2) and a=-9.8
v=u+at
0=sqrt(4.2)-9.8t
t=0.209s

Before the string is cut, P movies with SHM. So:
x=0.2, A=0.5, w=sqrt(20)
x=Acos(wt)
0.2=0.5cos(sqrt(20)t)
t=0.259s

t=0.259+0.209=0.468s

I assumed that you already had all the values I used. If you want, I can show you how I got them.
5. (Original post by dvs)
When the string is cut, P is a projectile. So consider vertical motion:
v=0, u=sqrt(4.2) and a=-9.8
v=u+at
0=sqrt(4.2)-9.8t
t=0.209s

Before the string is cut, P movies with SHM. So:
x=0.2, A=0.5, w=sqrt(20)
x=Acos(wt)
0.2=0.5cos(sqrt(20)t)
t=0.259s

t=0.259+0.209=0.468s

I assumed that you already had all the values I used. If you want, I can show you how I got them.
i always misread the f****ing questiob. i go the 0.209s answer but didnt realise i had to do time before string cut also :'(

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