A block of mass m1 = 3.0 kg slides along a frictionless table with a velocity of

Question

A block of mass m1 = 3.0 kg slides along a frictionless table with a velocity of +10 m/s. Directly in front of it, and moving with a velocity of +3.0 m/s, is a block of mass m2 = 5.0 kg. A massless spring with spring constant k = 1120 N/m is attached to the second block as in the figure below.(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 ?x. What is the value of ?x? (c) The blocks will eventually separate again. What are the velocities of the two blocks measured in the reference frame of the table, after they separate?

Explanation / Answer

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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