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 = 9.0 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?

m/s

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

cm

(c) The blocks will eventually separate again. What is the final velocity of each block measured in the reference frame of the table?

m/s (for m1)

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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