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thanks i worked it out from what you told me
the problem was i thought the velocity of q was 2u and not -2u. i thought they were travelling in the same directions not the opposite.
A smooth sphere P, of mass m kg, moving with speed 5u m/s collides directly with a smooth sphere Q, of mass 2m kg, moving towards each other with speed 2u m/s.
The coefficient of restitution between the spheres is e.
find the condition satisfied by e if the 2 spheres move in the same direction after the collision.
using the 2 normal equations i got their final speeds to be 3u-2ue and 3u-ue
because they move in the same direction i let 3u-2ue >0 and 3u-ue>0
solving these i get e<3 and e>3/2
the answer at the back of the book is e<1/14
i can go though my answer in more detail if you need it
thanks
i am not sure how your get equation 3 and 4.
using your letters:
the second equation is bkm+wm=ckm i thought and not the third equation?
now when i try to get the third equation i use the formula v1-v2/u1-u2 =-e so a-x/y-0 =-1/k
for the fourth equation i get c=b. i am not sure where you get b-w=c
when i solve these i get w=0
now to prove the sphere of mass km is moving perpendicular
i can rearrange a-x/y-0 =-1/k to ka-kx=-y now when i compare this formula with equation 1 they should be similar. the only way to make them similar is if a=0
so thats the first part of the question.
for the second part i am still confused
here is another similar question
Two smooth spheres of masses 4 kg and 2 kg impinge obliquely.
The 2 kg mass is brought to rest by the impact.
(i) Prove that, before impact, they were moving in directions perpendicular to each other.
(ii) Show that, as a result of impact, the kinetic energy gained by the 4 kg mass is equal to half that lost by the 2 kg mass.
the worked answer to the question is this
http://thephysicsteacher.ie/Exam%20Material/AppliedMaths/Scans/5.%20Collisions/1986%20b.pdf
for some reason in the second part of the question he only takes into consideration the i direction which does not make sense to me
Ok i think i have the four equations. I will post them
Kma+mx=ykm
Kmb=kmb
a-x/y =-1/k
B=b
A stands for the initial velocity of the km sphere in the i direction. B stands for the initial and final velocity of the sphere km in the j direction. Y stands for the final velocity of the sphere km in the i direction. X stands for the initial velocity of the m sphere in the x direction.
These are not right though are they?
are you sure.
i was thought the coefficient of friction always has to be negative
the second part of the question is show that as a result of the collision the kinetic energy lost by the sphere of mass m is k times the kinetic energy gained by the sphere of mass km.
the kinetic energy lost by mass m sphere is .5mx^2
kinetic energy gained by mass km sphere is .5(km)(y)^2-.5(km)(a)^2
i have used conservation of momentum in only the i direction as its pointless using it in the j direction.
a(km)+x(m)=y(km)+0(m)
then using the coefficient or restitution formula y-o/a-x =-1/k
solving these k=-1
actually wait i think i have it.
so when we compare these equations -a+x=yk and ak+x=yk they are inconsistent unless a=0 that means initially there speeds are bj for the sphere of mass km and xi for the sphere of mass m
Is this right??
Two smooth spheres of masses km and m collide obliquely. the sphere of mass m is brought to rest by the impact. the coefficient of restitution for the collision is 1/k (k greater or equal to 1) Prove before the impact the spheres were moving perpendicular to each other.
i have worked out k=-1. I know the sphere of mass m was xi+0j before and 0i+0j after.
i was wondering how to work it out but i got it. its 9*9*8
yes so 102,103,104 etc would be counted
how many integers between 100 and 1000 have different digits.?
im trying to solve this using permutations.
my textbook.
i can prove that they were moving in perpendicular directions before the impact . so the velocity of the 2kg after the impact is 0i +0j . the collision is along the i axis, so the j component does not change. this means that before the collision the 2kg was moving along with say pi+oj. this means that the 4kg must of been moving with 0i+xj and if they were moving perpindicular. i know this doesnt really prove they were moving before but ill use this information in working out there velocities
so 2p+4(0)=4y and y-o/0-p =-1/2 these equations dont work either
two smooth spheres of masses 2kg and 4kg collide obliquely . the 2 kg mass is brought to rest. the coefficient of restitution is 1/2
prove before the impact the spheres were moving in perpendicular directions to each other and find their velocities before and after the collision.
using conservation of momentum. 4x+2y=4p+2(0) where x and y are velocities before impact. p is velocity after impact of 4kg
the velocity in the j direction does not change as its along the i axis
newtons experimental law p-0/x-y =-1/2 dont know what to do next
i actually worked it out using dv/dt=dv/dh*dh/dt the answer i got was pi/4.im stuck on the next part though. what is the rate at which the free surface area of the water is increasing when the cone is 1/8th full
0
a vessel in the shape of a cone is standing on its apex. water flows in at a steady rate, of 1m^3 per minute. the vessel has a height of 2m and a diameter of 2m when the vessel is 1/8 full find the rate at which water is rising
v=1/3 pi r^2 h however i cant differentiate this
the probability of getting a head on flipping a biased coin is p. the coin is flipped n times producing a sequence containing m heads and (n-m) tails what is the probability of obtaining this sequence from n flips.
i cant understand the wording
i dont know what m2 means
however at the back of the book it says the answer is 4/3 . im confused
at the end you write v/2=√(s1+s2) ⇒v/2=v/√(3a) should it not be v/2=√(s1+s2)/√2 which would be v/2=√(2v^2/3a)/√2 when i solve this i dont get a=4/3
a particle is projected vertically upwards from a point P. at the same instant a second particle is let fall from rest vertically at q. q is directly above p. the 2 particles collide at a point r after t seconds . when the 2 particles collide they are travelling at equal speeds. prove that |pr|=3|rq|
i am trying to solve this with uvast equations for the first particle i have v=v s=r a=-g t=t for the second particle i have v=v s=p-r a=g t=t dont know where to go from here
a train accelerates uniformly from rest to a speed v m/s with uniform acceleration a m/s^2 it then decelerates uniformly to rest with uniform retardation 3a m/s^2. the total distance travelled is s metres. if the average speed for the whole journey is square root (s/2) find the value of a. does anyone know how to work this out using uvast equations
that is the right answer but how did you do it??
a particle passes point P with speed 10 m/s and moves in a straight line with uniform acceleration to a point Q . in the first second of motion after passing P it travels 14 m. in the last 5 seconds of motion before reaching Q it travelled 7/10 of |PQ|. find the acceleration of the particle and the distance from P to Q. i found out the acceleration by using uvast equations. s=ut +.5at^2 14=10(1)+.5(a)(1)^2 a=8 . however i dont know how to find the distance
Two particles A and B each of mass m are connected by a light inextensible string passing over a light, smooth, fixed pulley. Particle A rests on a rough plane inclined at α to the horizontal, where tan α =5/12 . Particle B hangs vertically 1 m above the ground. The coefficient of friction between A and the inclined plane is .5
The system is released from rest.
(ii) How far will A travel after B strikes the ground?
i know how to work out most of this question. the only thing i am confused on is what is the acceleration of mass of A when B strikes the ground. i would of said it was g/13 but it is -11g/13 . all i need to know is the acceleration of B.