In the figure, block 2 of mass 2.10 kg oscillates on the end of a spring in SHM

Question

In the figure, block 2 of mass 2.10 kg oscillates on the end of a spring in SHM with a period of 26.00 ms. The position of the block is given by x = (0.900 cm) cos(omega t + x/2). Block 1 mass 4.20 kg slides toward block 2 with a velocity of magnitude 9.00 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 6.50 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision? Number Units In the figure, two blocks (m = 2.50 kg and M = 10.5 kg) and a spring (k = 210 N/m) are arranged on a horizontal frictionless surface. The coefficient of static friction between the two blocks is 0.440. What amplitude of simple harmonic motion of the spring-blocks system puts the smaller block on the verge of slipping over the larger block? Number Units

Explanation / Answer

Solution: The external force required to put the smaller block on the verge of slipping over the larger block is = ?mg = 10.78 N. Therefore, the required condition of just slipping is (M+m)?2A = 10.78; here ?2 = k/(M+m). So, A = 10.78/k = 0.05 m.

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