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Incredible M2 Question!

A particle P of mass 0.2kg is at rest 3m from the edge of a horizontal bench. P receives an impulse of magnitude 4 NN s.

The bench is rough and the coefficient of friction between P and the bench is 2/5.

(c) the speed of P when P reaches the edge of the bench.


How?
bump :frown:
Reply 2
Avatar for Smeh
Smeh
14
robinyourpersie
A particle P of mass 0.2kg is at rest 3m from the edge of a horizontal bench. P receives an impulse of magnitude 4 NN s.

The bench is rough and the coefficient of friction between P and the bench is 2/5.

(c) the speed of P when P reaches the edge of the bench.


How?

Impulse equals change in momentum...
So 4=m(v-u)
20=v-0
so v=20.
Fmax=mur
fmax=0.4x0.2g
=0.784

F=ma
-0.784=0.2a
a=-3.92 (decelerating)

v^2=u^s+2as
v^2=400-(2x3.92x3)
v=19.4?

It's probably wrong though...Am I even vaguely headed in the right direction?
Reply 3
Avatar for Smeh
Smeh
14
.....
Smeh
Impulse equals change in momentum...
So 4=m(v-u)
20=v-0
so v=20.
Fmax=mur
fmax=0.4x0.2g
=0.784

F=ma
-0.784=0.2a
a=-3.92 (decelerating)

v^2=u^s+2as
v^2=400-(2x3.92x3)
v=19.4?

It's probably wrong though...Am I even vaguely headed in the right direction?


Yes your'e right You beautiful smeh!

I didn't know what to do once i had 3.92 which seems stupid now.
Reply 5
Avatar for Smeh
Smeh
14
robinyourpersie
Yes your'e right You beautiful smeh!

I didn't know what to do once i had 3.92 which seems stupid now.

:woo:
Happy cause I usually suck at M2 :biggrin:
Reply 6
Avatar for Mitch92uK
Mitch92uK
2
Impulse = change in momentum

Particle = at rest so u = 0

I = mv - mu
4 = 0.2 x v
v = 4/0.2 = 20.

Its then deccelerating. F = u R = 2/5 x 0.2 x 9.8 = 0.784
-0.784 = 0.2a
a = -0.784/0.2 = -3.92 ms^-2

v^2 = 20^2 + 2as
v^2 = 20^2 + (2x-3.92x3)
v^2 = 376.48
v = 19.40309254 ...
19.4ms^-1

Edit : I agree with Smeh!!
Smeh
.....


I'm confused.

Once I had 3.92 I used v=u+at which didn't work?

I then tried something else foing down the work done against friction route.

why didn't v=u+at work!?

v=20ms-1-3.92(3)=

Ok I've realised why. Because i'm an idiot.

I need a rest i've been doing too much M2.

and time and distance aren't the same are they lol?
Reply 8
Avatar for Smeh
Smeh
14
Mitch92uK
Impulse = change in momentum

Particle = at rest so u = 0

I = mv - mu
4 = 0.2 x v
v = 4/0.2 = 20.

Its then deccelerating. F = u R = 2/5 x 0.2 x 9.8 = 0.784
-0.784 = 0.2a
a = -0.784/0.2 = -3.92 ms^-2

v^2 = 20^2 + 2as
v^2 = 20^2 + (2x-3.92x3)
v^2 = 376.48
v = 19.40309254 ...
19.4ms^-1

Edit : I agree with Smeh!!

We even did it the same way :biggrin:
It took me ages to type that out I would've been pissed if someone had got there before me lol.
Reply 9
Avatar for Smeh
Smeh
14
robinyourpersie
I'm confused.

Once I had 3.92 I used v=u+at which didn't work?

I then tried something else foing down the work done against friction route.

why didn't v=u+at work!?

v=20ms-1-3.92(3)=

...because you don't have the time it took.
What's the point in wokring that out then using something like s=vt -0.5at^2 when you;ve already got a handy little equation like v^2 yadda yadda
Reply 10
Avatar for Mitch92uK
Mitch92uK
2
robinyourpersie
I'm confused.

Once I had 3.92 I used v=u+at which didn't work?

I then tried something else foing down the work done against friction route.

why didn't v=u+at work!?

v=20ms-1-3.92(3)=


You don't know/need to know the time taken + acceleration = -3.92, its slowing down.
Smeh
...because you don't have the time it took.
What's the point in wokring that out then using something like s=vt -0.5at^2 when you;ve already got a handy little equation like v^2 yadda yadda


Mitch92uK
You don't know/need to know the time taken + acceleration = -3.92, its slowing down.


Lol yeh, I just realised that see edit.

I think I need a lie down and a scotch. Cheers for help though both of you

<3

Your brain goes funny with too much M2 in one day.

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