All Mechanics DOUBTS HERE!!!

[halosh92]

http://www.scribd.com/doc/29694750/Edexcel-GCE-January-2010-Mechanics-M1-Marking-Scheme

thankyouuuuuuuuu

[Phosu]

jan 1010
7) d)

[vanibharutham]

the displacement vector from L to S is given by:

Position of S - Position of L

=> [(3t+9)i + (4t-6)j] - [ 18i + 6j ]

LS = (3t-9)i + (4t-12)j

therefore, the time from when DISTANCE, (not displacement), between L and S is 10 km is given by:

LS² = (3t - 9)² + (4t - 12)²
(10)² = 25t² - 150t + 225
0 = 25t² - 150t + 125
t² - 6t + 5 = 0

t = 5 and t = 1

Therefore,

T = 5 or T = 1

[halosh92]

jan 2006 -edexcel
Q7c- SMALL DOUBT
could someone just explain why is the direction reversed??
THX

coudl someone solve this plz!

[halosh92]

A particle A of mass 2 kg is moving along a straight horizontal line with speed 12 m s–1.
Another particle B of mass m kg is moving along the same straight line, in the opposite
direction to A, with speed 8 m s–1. The particles collide. The direction of motion of A
is unchanged by the collision. Immediately after the collision, A is moving with speed
3 m s–1 and B is moving with speed 4 m s–1. Find
(a) the magnitude of the impulse exerted by B on A in the collision
(b) the value of m.

i keep getting the impluse as negative
i took this -------------> as (+)  and this <---------- as (-)

[halosh92]

does anyone have june 2009 edexcel M1 ms ?
thx

[astarmathsandphysics]

A particle A of mass 2 kg is moving along a straight horizontal line with speed 12 m s–1.
Another particle B of mass m kg is moving along the same straight line, in the opposite
direction to A, with speed 8 m s–1. The particles collide. The direction of motion of A
is unchanged by the collision. Immediately after the collision, A is moving with speed
3 m s–1 and B is moving with speed 4 m s–1. Find
(a) the magnitude of the impulse exerted by B on A in the collision
(b) the value of m.

i keep getting the impluse as negative
i took this -------------> as (+)  and this <---------- as (-)
conservation of momentum
12*2-8m=3*2+4m
24-8m=6+4m
24-6=8m+4m
18=12m so m=1.5
impulse=mv-mu for A
2*3-2*12=-18

[astarmathsandphysics]

june 2006 7c
r(t)=start point +t*velocity
=16i+5j+3*(-2.5i+6j)=8.5i+23j

[halosh92]

A particle A of mass 2 kg is moving along a straight horizontal line with speed 12 m s–1.
Another particle B of mass m kg is moving along the same straight line, in the opposite
direction to A, with speed 8 m s–1. The particles collide. The direction of motion of A
is unchanged by the collision. Immediately after the collision, A is moving with speed
3 m s–1 and B is moving with speed 4 m s–1. Find
(a) the magnitude of the impulse exerted by B on A in the collision
(b) the value of m.

i keep getting the impluse as negative
i took this -------------> as (+)  and this <---------- as (-)
conservation of momentum
12*2-8m=3*2+4m
24-8m=6+4m
24-6=8m+4m
18=12m so m=1.5
impulse=mv-mu for A
2*3-2*12=-18

sir this is my problem...the impulse is 18 not -18 :SS am always getting it negative in all papers. :s

[nid404]

Impulse is a scalar quantity...i think 

[halosh92]

Two particles A and B are moving on a smooth horizontal plane. The mass of A is 2m and
the mass of B is m. The particles are moving along the same straight line but in opposite
directions and they collide directly. Immediately before they collide the speed of A is 2u
and the speed of B is 3u. The magnitude of the impulse received by each particle in the
collision is 7mu/2

Find
(a) the speed of A immediately after the collision,

(b) the speed of B immediately after the collision

[astarmathsandphysics]

By B on A - I read it the other way round
mV-mu for B
1.5*4--1.5*8=18

[7ooD]

can someone plz explain question 5 part c and d in 2001 june i dnt get wat the markscheme says qp and markschemes are attached

[astarmathsandphysics]

here

[Muho]

Someone help me with january 2009 Q7 (b)..

I have the mark scheme i just don't understand how they freakn get there =/