Two carts of equal mass, m = 0.250 kg, are placed on a frictionless track that h

Question

Two carts of equal mass, m = 0.250 kg, are placed on a frictionless track that has a light spring of force constant k = 41.0 N/m attached to one end of it, as in Figure P6.60. The red cart is given an initial velocity of 0 = 3.35 m/s to the right, and the blue cart is initially at rest. Figure P6.60 (a) If the carts collide elastically, find the velocity of the carts just after the first collision. red cart m/s blue cart m/s (b) If the carts collide elastically, find the maximum compression in the spring. m

Explanation / Answer

There is conservation of momentum: m1*v0 = (m1v1+m2v2); m1 = m2, so m*v0 = m(v1+v2), therefore v1+v2 = v; The kinetic energy before collision is .5m*v0^2; the kinetic energy after collision is .5*m*v1^2 + .5*m*v2^2, so v0^2 = v1^2 + v2^2. v1 = v0 - v2 v0^2 = (v0-v2)^2 + v2^2 v0^2 =v0^2 - 2v0*v2 + v2^2 +v2^2 0 =-2*v0*v2 + 2*v2^2 0 = -v0 + v2 v2 = v0 then v1 = 0 EDIT: Actually this has two solutions, v2=v0 or v2=0; If v2=0 then v1=v0, meaning cart 1 moves through cart 2, which cannot happen. The energy of the second cart is .5*m*v0^2. The energy in compressing the spring is ?F*dx = ?kx*dx = (k/2)*s^2 where s is the spring compression. Therefore (k/2)*s^2 = .5*m*v0^2 s^2 = (2/k)*.5*m*v0^2 s = v0*v(m/k)

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