Author Topic: M.E.C.H.A.N.I.C.S-edexcel doubts! :-[  (Read 873 times)

Offline halosh92

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M.E.C.H.A.N.I.C.S-edexcel doubts! :-[
« on: May 19, 2010, 06:57:41 pm »
jan 2007
Q7f
plzz someone explain :S
thx
everyday we wake up is a miracle, then how do we say miracles dont happen?????

Offline cooldude

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Re: M.E.C.H.A.N.I.C.S-edexcel doubts! :-[
« Reply #1 on: May 19, 2010, 07:15:21 pm »
jan 2007
Q7f
plzz someone explain :S
thx

okay for this basically u find out the time the for P to come to rest and then double that value---.
as the string is now slack for this period there is no tension in the string so the deceleration of P will be-->
F=ma, -30sin30=3a
a=-10sin30=-5
the time taken for Q to reach the ground is-->
0.8=0.5*0.98*t^2
t=1.28 sec
therefore final speed of P=
v=0+(1.28*0.98)
v=1.24 m/s
0=1.24-5t
t=0.248 sec
this is the time P takes to come to rest therefore the time for it to become taut again will be double that=0.248*2=0.496 sec, By the way there might be some inaccuracies in the final answer as i did not use exact values
we double the time because wen P reaches rest, Q is still incapable of movement as the string is still slack, but then when P goes back down the plane and when it reaches the point where it initially was i.e. 0.51 sec previously, the string becomes taut again and Q rises again, we double the time because we consider both the period when P travels up the plane and back down. P moves down the plane because of its weight, there is no other force acting on P as the tension in the string is non-existent for the duration of the time the string is slack, i.e. the interval of 0.51 sec, hope u get it.

Offline halosh92

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Re: M.E.C.H.A.N.I.C.S-edexcel doubts! :-[
« Reply #2 on: May 20, 2010, 06:03:43 pm »
thakyouuuuuuuuu
i have one more question:
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 (-)

and how do i knw the direction of B after the collision?

« Last Edit: May 20, 2010, 06:06:09 pm by halosh92 »
everyday we wake up is a miracle, then how do we say miracles dont happen?????