••36. In Fig. 15-40 ,block 2 of mass 2.0 kg oscillates on the end of a spring in

Question

••36.   In Fig. 15-40,block 2 of mass 2.0 kg oscillates on the end of a spring in SHMwith a period of 20 ms. The position of the block is givenby .Block 1 of mass 4.0 kg slides toward block 2 with a velocity ofmagnitude 6.0 m/s, directed along the spring’s length. Thetwo blocks undergo a completely inelastic collision attime .(The duration of the collision is much less than the period ofmotion.) What is the amplitude of the SHM after the collision?    FIGURE 15-40    Problem 36.





I keep getting a different answer than what my professor indicatedshould be right. He gives us the answers to some of them so we cancheck to see if we did it right, and his answer for the A =2.42E-2m but i keep getting different answers each time I doit. In Fig. 15-40,block 2 of mass 2.0 kg oscillates on the end of a spring in SHM with a period of 20 ms. The position of the block is given by .Block 1 of mass 4.0 kg slides toward block 2 with a velocity of magnitude 6.0 m/s, directed along the spring?s length. The two blocks undergo a completely inelastic collision at time .(The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision? I keep getting a different answer than what my professor indicated should be right. He gives us the answers to some of them so we can check to see if we did it right, and his answer for the A =2.42E-2m but i keep getting different answers each time I do it.

Explanation / Answer

Given that          m1 = 4.0 kg          m2 = 2.0kg          T = 20 ms           v1 = 6m/s The spring constant k = 42 m1 /T2                                   = 4 * 3.14 * 3.14 * 4 / (20 x 10-3)2                                   = 1.97 x 105 N / m At collision t = 2t / T = / 2 so that x = -xm If the spring is stretched at this moment is 1 cm we have         xm = - 0.01m From conservation of momentum we have         m1v1 = m v      ==> v = 4 * 6 / 6 = 4 m/s Kinetic energy of the system after collision we have        KE = 0.5mv2 =0.5*6*4*4 = 48J and PE = kx*x = 1.97 x 105 * 0.01*0.01 = 10 J Total mechanical energy = kx2      ==>   kx2 = 58J      ==> x = 58 / 1.97 x105               = 0.024 m = 2.4 x 10-2 m

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