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quick question about coeffecient of restitution

A smooth sphere P of mass m is moving in a straight line with speed U on a smooth horizontal table. Another smooth sphere Q of mass 2m is at rest on the table. The sphere P collides directly with Q. After the collision the direction of motion of P is unchanged. The spheres have the same radii and the coefficient of restitution between P and Q is e. By modelling the spheres as particles,

(a) show that the speed of Q immediately after the collision is 1/3(1 + e)U
Which ive done.

(b) find the range of possible values of e. but im not too sure how u do this?

any ideas please?
find the speed of P and make it greater than 0

so e = (v2-v1)÷u
v2= v1 + eu
mu= mv1 + 2mv2

u = 3v1 + 2eu
1/3(1-2e)u = v1

v1 > 0 so
1/3 - 2/3e > 0
e < 1/2
Reply 2
but then i get e=1/2u
sry, edited above

v1 must be greater than zero cos it says it continues to move in the same direction
Reply 4
oh ok i get it now, thanks. one more question:

A rocket R of mass 100 kg is projected from a point A with speed 80ms^-1 at an angle of elevation of 60°. The point A is 20 m vertically above a point O which is on horizontal ground. The rocket R moves freely under gravity. At B the velocity of R is horizontal. By modelling R as a particle,

find (a) the height in m of B above the ground,

(b) the time taken for R to reach B from A.

When R is at B, there is an internal explosion and R breaks into two parts P and Q of masses 60 kg and 40 kg respectively. Immediately after the explosion the velocity of P is 80ms^-1 horizontally away from A. After the explosion the paths of P and Q remain in the plane OAB. Part Q strikes the ground at C. By modelling P and Q as particles,

(c) show that the speed of Q immediately after the explosion is 20ms^-1

(d) find the distance OC.

thanks a lot.
Freeway
A smooth sphere P of mass m is moving in a straight line with speed U on a smooth horizontal table. Another smooth sphere Q of mass 2m is at rest on the table. The sphere P collides directly with Q. After the collision the direction of motion of P is unchanged. The spheres have the same radii and the coefficient of restitution between P and Q is e. By modelling the spheres as particles,

(a) show that the speed of Q immediately after the collision is 1/3(1 + e)U
Which ive done.

(b) find the range of possible values of e. but im not too sure how u do this?

any ideas please?


a) Let the direction of P be positive.
Coefficient of restitution
e = (Vq - Vp)/U
Vq - Vp = eU
Vq = eU + Vp (1)

Conservation of momentum
mU + 0 = mVp + 2mVq
U = Vp + 2Vq
Vp = U - 2Vq (2)

(2) into (1)
Vq = eU + U - 2Vq
3Vq = U(e+1)
Vq = U(e+1)/3

Vp = Vq - eU
Vp = eU/3 + U/3 - eU
Vp = -2eU/3 + U/3
Vp = U/3 - 2eU/3
Vp = U(1-2e)/3

b)
The direction of P is unchanged, therefore, Vp > 0
U(1-2e)/3 > 0
1 - 2e > 0
1 > 2e
2e < 1
e < ½
Freeway
oh ok i get it now, thanks. one more question:

A rocket R of mass 100 kg is projected from a point A with speed 80ms^-1 at an angle of elevation of 60°. The point A is 20 m vertically above a point O which is on horizontal ground. The rocket R moves freely under gravity. At B the velocity of R is horizontal. By modelling R as a particle,

find (a) the height in m of B above the ground,

(b) the time taken for R to reach B from A.

When R is at B, there is an internal explosion and R breaks into two parts P and Q of masses 60 kg and 40 kg respectively. Immediately after the explosion the velocity of P is 80ms^-1 horizontally away from A. After the explosion the paths of P and Q remain in the plane OAB. Part Q strikes the ground at C. By modelling P and Q as particles,

(c) show that the speed of Q immediately after the explosion is 20ms^-1

(d) find the distance OC.

thanks a lot.


a) horizontal u=80cos60
a=0


vertical u=80sin60
a=-9.8
v=0 (because velocity is horizontal)
v²=u² + 2as
0= 4800 + 2(-9.8)(s)
s&#8776; 245
total distance above ground = 245 + 20 = 265

b) t= (v-u)/a = (0-80sin60)/-9.8 = 7.07 s

c) for this one use conservation of momentum

100.(80cos60) = 60.80 - 40.v2
v2 = 20ms^-1

d) first to find OA's horizontal displacement (same as AB's)
horizontal u=80cos60
t= 7.07
a=0
s= ut + ½at² = 40(7.07) + ½(0)(7.07²) &#8776; 283

now to find AC, use the projectile of Q
vertical u=0
a=9.8
s= 265
s= ut + ½at²
265= 0 + ½(9.8)t²
t= 7.35

horizontal u=20
a=0
t=7.35
s= ut + ½at² = 20(7.35) + 0 = 147

Therefore, OA - AC = OC
283-147 = 136
Reply 7
cheers featherflare, i owe you one :biggrin:
No problem. Happy to help :smile: