More M1 Help Needed... Please!

Got some more questions I wanna check/ learn how to do. Any help would be great.

Thanks again, Adam.

1. See attachment below please... Two particles A and B have masses 3m and km respectively, where k>3. They are connected by a light inextensible string which passes over a smooth fixed pulley. The system is released from rest with the string taut and the hanging parts of the string vertical. While the particles are moving freely, A has an acceleration of magnitude 2/5 g.

a) Find, in terms of m and g, the tension in the string.

b) State why B also has an acceleration of 2/5 g.

c) Find the value of k.

d) State how you have used the fact that the string is light.

_________________________________________________________________

2. A particle P moves in a straight line with constant velocity. Initially P is at the point A with position vector (2i-j)m relative to a fixed origin O, and 2 seconds later it is at the point B with position vector (6i+j)m.

a) Find the velocity of P.

b) Find, in degrees to 1 d.p., the size of the angle between the direction of motion of P and the vector i.

Three seconds after it passes B the particle P reaches the point C.

c) Find, in m to 1 d.p., the distance OC.

_________________________________________________________________

3. A sledges of mass 78kg is pulled up a slope by means of a rope. The slope is modelled as a rough plane inclined at an angle x to the horizontal, where tan x=5/12. The rope is modelled as light and inextensible and is in a line of greatest slope of the plane. The coefficient of friction between the sledge and the slope is 0.25. Given that the sledge is accelerating up the slope with acceleration 0.5ms^2,

a) Find the tension in the rope.

The rope suddenly breaks. Subsequently the sledge comes to an instantaneous rest and then starts sliding down the slope.

b) Find the acceleration of the sledge *after* it has come to instantaneous rest.

Doc12.doc23.6KB

Reply 1

Fermat

a=(2/5)g = 0.4g

a)
forces on A
==========

(T-3mg)=(3m)a
(T-3mg)=3m*0.4g
T-3mg=1.2mg
T=4.2mg
======

b)
Since the connection between them is under tension and inextensible, then their movement is the same.

c)
forces on B
=========

(kmg - T)=(km)*a
(kmg - 4.2mg) = km*0.4g
kmg - 4.2mg = 0.4kmg
k - 4.2 = 0.4k
0.6k=4.2
k=7
===

d)
Since the string is light, then its mass is ignored otherwise some of T would have been needed to provide an accelerating force on it.

Reply 2

Fermat

2)
A=(2i - j) m
B=(6i + j) m, (after 2s)

a)
P moves from the poinr A to the point B.
Therefore its velocity is parallel to the vector AB

& AB = B-A
AB=(6i+j) - (2i-j)
AB=4i+2j, travelled in 2s

therefore velocity,

v=AB/2
v=2i+j m/s
======

b)
P=(2.i + (-1).j)
θ=tan^(-1)(-1/2)
θ=tan(-0.5)
θ=-26.6°
=====

c)
c=A+3v
C=(2i-j)+3(2i+j)
C=8i+2j
=====

|C|=√(8² + 2²)
|C|=√(64+4)
|C|=√68
|C|=8.2
=====

Reply 3

Fermat

3)
µ=0.25, a=0.5 m/s², M=78 kg

tan x = 5/12
sin x = 5/13
cos x = 12/13

a)
Forces down slope, Fr & Mg.sin x
===========
Fr=µMg.cos x = 0.25*78g*(12/13)
Fr=176.58 N
=======

Mg.sin x = 78g*(5/13)
Mg.sin x = 294.3 N
===========

Force up slope, T , tension in the rope
=========

Accelerating force is,

F=T - Fr - Mg.sin x
F = T - 176.58 - 294.3
F = T - 470.88 N
==========

F=ma
===

T-470.88 = 78g*0.5
T-470.88 = 382.59
T=1053.47 N
=======

b)
After stopping, the accelerating force down the slope is,

F=Mg.sin x - Fr
F=294.3 - 176.58
F=117.72 N
======

F=ma
===

117.72=87g.a
a=117.72/765.18
a=0.1538 m/s²
========

Reply 4

adk

Thankyou SO much. I really appreciate it.

I'll print that off tomorrow morning, and then set about trying to understand it.

Cheers.

Adam

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