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

Question

A block of mass m1 = 2.0 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) Before m1 runs into the spring, what is the velocity of the center of mass of the system?
1 m/s

(b) After the collision, the spring is compressed by a maximum amount ?x. What is the value of ?x?
2 cm

(c) The blocks will eventually separate again. What is the final velocity of each block measured in the reference frame of the table?
3 m/s (for m1)
4 m/s (for m2)

Explanation / Answer

(a) Velocity of CM = m1v1 + m2v2/(m1+m2) = 61/11 = 5.55 m/s (b) Spring is compressed maximum when velocities are equal Let it be v by conservation of momentum, m1v1 + m2v2 = m1v + m2v v = 5.55 m/s Total energy of the system remains the same So, (1/2)m1v1^2 + (1/2)m2v2^2 = (1/2)(m1+m2)v^2 +(1/2)kx^2 (1/2)kx^2 = 62.36 J x = 33.37 cm (c) Final velocities in elastic collision are vf1 = (v1(m1-m2) + 2*m2*v2)/(m1+m2) = 1.09 m/s vf2 = (v2(m2-m1) + 2*m1*v1)/(m1+m2) = 8.09 m/s

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