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M2 question
6. A particle P of mass 3m is moving with speed 2u in a straight line on a smooth horizontal table. The particle P collides with a particle Q of mass 2m moving with speed u in the opposite direction to P. The coefficient of restitution between P and Q is e.
(a) Show that the speed of Q after the collision is u/5(9e + 4).
This parts fine.
As a result of the collision, the direction of motion of P is reversed.
(b) Find the range of possible values of e.
I try two different methods and get e<2/3 and e>1/3???
Given that the magnitude of the impulse of P on Q is 32mu/5
(c) find the value of e.
I = mv - mu
32mu/5 = 2m[u/5(9e + 4) - (-u)]
32mu =10m[u/5(9e + 4) + u)]
32mu = 2mu(9e + 4) + 10mu
22mu = 18mue + 8mu
14mu = 18mue
e = 14mu/18mu = 7/9
Yep, that works.
(a) Show that the speed of Q after the collision is u/5(9e + 4).
This parts fine.
As a result of the collision, the direction of motion of P is reversed.
(b) Find the range of possible values of e.
I try two different methods and get e<2/3 and e>1/3???
Given that the magnitude of the impulse of P on Q is 32mu/5
(c) find the value of e.
I = mv - mu
32mu/5 = 2m[u/5(9e + 4) - (-u)]
32mu =10m[u/5(9e + 4) + u)]
32mu = 2mu(9e + 4) + 10mu
22mu = 18mue + 8mu
14mu = 18mue
e = 14mu/18mu = 7/9
Yep, that works.
Widowmaker
6. A particle P of mass 3m is moving with speed 2u in a straight line on a smooth horizontal table. The particle P collides with a particle Q of mass 2m moving with speed u in the opposite direction to P. The coefficient of restitution between P and Q is e.
(a) Show that the speed of Q after the collision is u/5(9e + 4).
This parts fine.
As a result of the collision, the direction of motion of P is reversed.
(b) Find the range of possible values of e.
I try two different methods and get e<2/3 and e>1/3???
Given that the magnitude of the impulse of P on Q is 532mu,
(c) find the value of e.
(a) Show that the speed of Q after the collision is u/5(9e + 4).
This parts fine.
As a result of the collision, the direction of motion of P is reversed.
(b) Find the range of possible values of e.
I try two different methods and get e<2/3 and e>1/3???
Given that the magnitude of the impulse of P on Q is 532mu,
(c) find the value of e.
taking initial direction of P as +ve:
a)
2u x 3m - u x 2m = 3m x Vp + 2m Vq
4u = 3Vp + 2Vq
Vp = [4u - 2Vq] / 3
e = (Vq - Vp)/ 3u
3eu = Vq - [4u - 2Vq]/3
9eu = 5Vq - 4u
Vq = u/5 x (9e+4) as required
b) Vp = [4u-2/5 u (9e+4) ] / 3
Vp = 4/5 - 6/5e
as Vp reverses direction ==> then Vp is -ve now:
Vp < 0
4/5 - 6/5e < 0
4 < 6e
e > 2/3
c) I = m(V-U)
32mu / 5 = (2m)(Vq - -u)
16u/5 = ( u/5 x (9e+4) + u)
16/5 = ( (9e+4)/5 + 1)
16/5= [9e+9]/5
e = 16/9 - 1 = 7/9
assuming that P and Q are both travelling in the same direction after the impact ( the same as P was before impact), then I get
V_q = u/5(9e + 4), the same as yourself.
Call this direction +ve.
After impact, the direction of P is reversed - the velocity of P is -ve.
V_q = u/5(9e + 4),
and you can work out
V_p = u/5(4 - 6e),
Since V_p is -ve,
u/5(4 - 6e) < 0
4 - 6e < 0
4 < 6e
(2/3) < e
e > (2/3)
=======
V_q = u/5(9e + 4), the same as yourself.
Call this direction +ve.
After impact, the direction of P is reversed - the velocity of P is -ve.
V_q = u/5(9e + 4),
and you can work out
V_p = u/5(4 - 6e),
Since V_p is -ve,
u/5(4 - 6e) < 0
4 - 6e < 0
4 < 6e
(2/3) < e
e > (2/3)
=======
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