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M2 Collisions
1) A smooth sphere A of mass m is moving with speed u on a smooth horizontal table when it collides directly with another smooth sphere B of mass 3m, which is at rest on the table. The coefficient of restitution between A and B is e. The spheres have the same radius and are modelled as particles.
a) Show that the speed of B immediately after the collision is (1+e)u/4.
b) Find the speed of A immediately after the collision.
Immediately after the collision, the total kinetic of the spheres is (1/6)mu^2.
c) Find the value of e.
d) Hence show that A is at rest after the collision.
a) Show that the speed of B immediately after the collision is (1+e)u/4.
b) Find the speed of A immediately after the collision.
Immediately after the collision, the total kinetic of the spheres is (1/6)mu^2.
c) Find the value of e.
d) Hence show that A is at rest after the collision.
Conservation of momentum:
mu = mvA + 3mvB
u = vA + 3vB (1)
Newton's law of restitution:
vB - vA = eu (2)
a)
(1) + (2) gives:
4vB = u+eu
=> vB = u(1+e)/4, as required.
b)
Substituting this value of vB in either (1) or (2) gives:
vA = u(1-3e)/4
c)
KE = 0.5m(vA)² + 0.5(3m)(vB)² = (1/6)mu²
(vA)² + 3(vB)² = (1/3)u²
[u(1-3e)/4]² + 3[u(1+e)/4]²= (1/3)u²
e² = 1/9
=> e = 1/3 (-1/3 is neglected because e >= 0).
d) vA = u(1-3e)/4 = u(1-1)/4 = 0, as required.
mu = mvA + 3mvB
u = vA + 3vB (1)
Newton's law of restitution:
vB - vA = eu (2)
a)
(1) + (2) gives:
4vB = u+eu
=> vB = u(1+e)/4, as required.
b)
Substituting this value of vB in either (1) or (2) gives:
vA = u(1-3e)/4
c)
KE = 0.5m(vA)² + 0.5(3m)(vB)² = (1/6)mu²
(vA)² + 3(vB)² = (1/3)u²
[u(1-3e)/4]² + 3[u(1+e)/4]²= (1/3)u²
e² = 1/9
=> e = 1/3 (-1/3 is neglected because e >= 0).
d) vA = u(1-3e)/4 = u(1-1)/4 = 0, as required.
->u ->0 ->v1 ->v2
m 3m COLLIDE m 3m
By conservation of momentum: mu=mv1 + 3mv2
Cancelling m u=v1 + 3v2
By Newton's Law of restitution: e= speed of separation of particles/
speed of approach of particles
e=v2-v1/ u
Rearranging: v1=v2 - eu
Substituting into the equation of momentum: u=v2-eu+ 3v2
u(1+e)/4 = v2
Substituting v2 into the equation gives: v1= u(1-3e)/4
The final kinetic energy (general formula being 1/2*mass*velocity^2):
1/2m * [1/4u (1-3e)]^2 + 3/2m * [1/4u(1+e)]^2
-> 1/2m * 1/16 u^2(1-3e)^2 + 3/2m * 1/16u^2 (1+e)^2
-> 1/32 mu^2 [(1-3e)^2 + 3(1+e)^2] which is equal to 1/6mu^2 as the
question tells you.
Therefore: 1/32 (1+9e^2 -6e+ 3(1+e^2+2e)) = 1/6
-> 1/32 (4+12 e^2) = 1/6
-> 1/8 + 3/8 e^2 = 1/6
-> 3/8 e^2= (4-3)/24 = 1/24
-> e^2 = 1/9
-> e= 1/3
It's really painstaking typing maths working out on computer, hope you get it!
m 3m COLLIDE m 3m
By conservation of momentum: mu=mv1 + 3mv2
Cancelling m u=v1 + 3v2
By Newton's Law of restitution: e= speed of separation of particles/
speed of approach of particles
e=v2-v1/ u
Rearranging: v1=v2 - eu
Substituting into the equation of momentum: u=v2-eu+ 3v2
u(1+e)/4 = v2
Substituting v2 into the equation gives: v1= u(1-3e)/4
The final kinetic energy (general formula being 1/2*mass*velocity^2):
1/2m * [1/4u (1-3e)]^2 + 3/2m * [1/4u(1+e)]^2
-> 1/2m * 1/16 u^2(1-3e)^2 + 3/2m * 1/16u^2 (1+e)^2
-> 1/32 mu^2 [(1-3e)^2 + 3(1+e)^2] which is equal to 1/6mu^2 as the
question tells you.
Therefore: 1/32 (1+9e^2 -6e+ 3(1+e^2+2e)) = 1/6
-> 1/32 (4+12 e^2) = 1/6
-> 1/8 + 3/8 e^2 = 1/6
-> 3/8 e^2= (4-3)/24 = 1/24
-> e^2 = 1/9
-> e= 1/3
It's really painstaking typing maths working out on computer, hope you get it!
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