A block of mass m1 = 2.0 kg slides along a frictionless table with a speed of 10

Question

A block of mass m1 = 2.0 kg slides along a frictionless table with a speed of 10 m/s. Directly in front of it, and moving in the same direction with a speed of 3.0 m/s, is a block of mass mi = 5.0 kg. A massless spring that has a force constant k = 1120 N/m is attached to the second block as in Figure 8-47. (a) What is the velocity of the center of mass of the system? (b) During the collision, the spring is compressed by a maximum amount Delta x. What is the value of Delta x? (c) The blocks will eventually separate again. What are the final velocities of the two blocks measured in the reference frame of the table, after they separate?

Explanation / Answer

I'm a bit rusty on these type of problems, but I'll do my best. Velocity of the center of mass of the system is the sum of themomentums divided by the total mass. vcom = (m1v1 +m2v2)/(m1+ m2) vcom = (2*10 + 5*3)/(2+5) vcom = 5 m/s At the point at which the spring is most compressed, the two blocksmust be moving at the same speed (if the first block is movingfaster, then it can compress the spring more; if the second blockis moving faster, then the two are moving away from each other).However, the momentum of the system must be the same. (m1vf +m2vf)/(m1+ m2) = 5 vf = 5 However, some energy will be transferred to the spring. Therefore,the difference in kinetic energy of the two blocks should tell ushow much the spring compresses. .5m1v12 +.5m2v22 -(.5m1vf2 +.5m2vf2) = .5kx^2 122.5 - 87.5 = 560x^2 x = .25 m Since this is an elastic collision, we can use the "magic"equations for elastic collisions. They are as follows: v1f=((m1-m2)/(m1+m2))(v1i)+(2m2/(m1+m2)v2i v2f =(2m1/(m1+m2))v1i +((m2-m1)/(m1+m2))v2i They evaluate to: v1f= 0 m/s v2f =7 m/s

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