A block of mass m1 = 1.6 kg slides along a frictionless table with a speed of 8

Question

A block of mass m1 = 1.6 kg slides along a frictionless table with a speed of 8 m/s. Directly in front of it, and moving in the same direction, is a block of mass m2 = 4.2 kg moving at 2.8 m/s. A massless spring with spring constant k = 1180 N/m is attached to the near side of m2, as shown in Fig. 10-35. 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.)
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Explanation / Answer

Mass of the first block, m1 = 1.6 kg Mass of the second block, m2 = 4.2 kg Initial velocity of the first block, v1 = 8 m/s Initial velocity of the second block, v2 = 2.8 m/s Spring constant, k = 1180 N/m In a perfectly inelastic collision, m1 v1 + m2 v2 = ( m1 + m2 ) v [ from the law of conservation of momentum ] Common velocity after collision, v = 1.6 * 8 + 4.2 * 2.8 / ( 1.6 + 4.2 )                                                     = 4.234 m/s Loss in kinetic energy = (1/2) m1 v1^2 + (1/2) m2 v2^2 - (1/2) (m1 + m2 ) v^2 = 0.5 * 1.6 * 8^2 + 0.5 * 4.2 * 2.8^2 - 0.5 * * 4.23^2 = 15.66 J Potential energy stored in the spring = Loss of kinetic energy (1/2) k x^2 = 15.66 0.5 * 1180 * x^2 = 15.66 x = 0.162 m

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