A block of mass m1 = 2.4 kg slides along a frictionless table with a speed of 10

Question

A block of mass m1 = 2.4 kg slides along a frictionless table with a speed of 10 m/s. Directly in front of it, and moving in the same direction, is a block of mass m2 = 4.4 kg moving at 2.8 m/s. A massless spring with spring constant k = 1100 N/m is attached to the near side of m2, as shown in Fig. 11-38. When the blocks collide, what is the maximum compression of the spring? (Hint: At the moment of maximum compression of the spring, the two blocks move as one. Find the velocity by noting that the collision is completely inelastic to this point.) __m

Explanation / Answer

at the moment of maximum compression the velocity of two block is same
since spring force is internal force to the two blocks system, applying conservation of linear momentum
m1*v1 + m2*v2 = (m1 + m2 )*v
so v = 5.3411 m/s
now applying conservation of energy since there are no dissipating forces
(1/2)*m1*v12 + (1/2)*m2*v22 = (1/2)*(m1 +m2)*v2 + (1/2)*k*x2

so we get x=27.0538cm

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