Three particles, A, B and C are spaced out along the x-axis from left-to-right.

Question

Three particles, A, B and C are spaced out along the x-axis from left-to-right. The masses of A, B and C are such that m_A = m_C = 3m_B. Particle A is given a positive x-velocity u_A, while the others start at rest with u_B = u_C = 0. A series of collisions then takes place. Denoting the velocities after collision number 1 by u', after collision number 2 by u'', etc, and assuming that the collisions are elastic, calculate:

(a) The velocities u'_A, u'_B, u'_C following the rst collision, in terms of u_A.
(b) The velocities u''_A, u''_B, u''_C following the second collision, also in terms of u_A.
(c) If instead all the collisions were completely inelastic, and the particles stuck together every time they collided, what would be their nal velocities after the rst two collisions?

Explanation / Answer

momentum before colliion=momentum after collision and kinetic energy before collision=kinetic energy after collision

m1u1+m2u2+m3u3=m1v1+m2v2+m3v3

1/2m1u12+1/2m2u22+1/2m3u32=1/2m1v12+1/2m2v22+1/2m3v32

(a)substituting m1(A)=3m1(B)=m3(C)

U1(A)=U1

U2(B)=U3(C)=0

V1(A)=V2(B)=V3(C)=U1

in the momentum equation we get,

u1(A)=u1 * 5/3A=u1 *5/3*B=u1 * 5/3 *c

(b)U1(A)=U1(B)=U1(C)=U1(A)

(C)final velocities becomes 5/3 times of initial velocities

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