In the figure below, block 2 of mass 4.0 kg oscillates on the end of a spring in

Question

In the figure below, block 2 of mass 4.0 kg oscillates on the end of a spring in SHM with a period of 20 ms. The position of the block is given by x = (1.0 cm) cos(omega t + pi / 2). Block 1 of mass 5.0 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 t = 5.0 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision? m

Explanation / Answer

K = 394784.1 then common velocity of both = { 5*6 + 4*3.14 }/ 9 = 4.729 m/s amplitude = x 0.5 * 394784.1 * x^2 = 0.5 * 9 * 4.729^2 amplitude = 0.203 m

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