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.
