A block of mass 3 kg slides along a frictionless table with a speed of 5 m/s. Di

Question

A block of mass 3 kg slides along a frictionless table with a speed of 5 m/s. Directly in front of it, and moving in the same direction, is a second block of mass 1.7 kg moving at 3.4 m/s. A massless spring with spring constant 401 N/m is attached to the near side of the second block. 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 at this point.)

Explanation / Answer

Mass of the first block, m1 = 3 kg Mass of the second block, m2 = 1.7 kg Initial velocity of the first block, v1 = 5 m/s Initial velocity of the second block, v2 = 3.4 m/s Spring constant, k = 401 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 = 3 * 5 + 1.7 * 3.4 / ( 3 + 1.7 )                                                     = 4.42 m/s Loss in kinetic energy = (1/2) m1 v1^2 + (1/2) m2 v2^2 - (1/2) (m1 + m2 ) v^2                                   = 0.5 * 3 * 5^2 + 0.5 * 1.7 * 3.4^2 - 0.5 * 4.7 * 4.42^2                                   = 37.5 + 9.83 - 45.94                                   = 1.39 J Potential energy stored in the spring = Loss of kinetic energy (1/2) k x^2 = 1.39 0.5 * 401 * x^2 = 36.1 x = 0.0833 m

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