Chapter 15, Problem 034 In the figure, block 2 of mass 2.50 kg oscillates on the

Question

Chapter 15, Problem 034 In the figure, block 2 of mass 2.50 kg oscillates on the end of a spring in SHM with a period of 12.00 ms. The position of the block is given by x = (0.900 cm) cos(Ohm t + pi/2). Block 1 of mass 5.00 kg slides toward block 2 with a velocity of magnitude 3.00 m/s, directed along the spring's length. The two blocks undergo a completely inelastic collision at time t = 3.00 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 The tolerance is +/-2%

Explanation / Answer

w = 2*pi/T = 2*3.14*1000/12 = 523.33 s^-1

k = m2*w^2 = 547755.55 N/m

at t = 3ms

x = 0.9*cos(523.33*0.003 + pi/2)

x = -0.9 cm

at t = 3ms the block 2 is at extreme point with speed v2 = 0

from momentumconservation m1*v1 + m2*v2 = (m1+m2)*V

V = ((5*3)+(2.5*0))/(5+2.5) = 2 m/s

KE = PE

0.5*(m1+m2)*V^2 = 0.5*K*A'^2

0.5*(2.5+5)*2*2 = 0.5*547755.55*A^2

A' = 0.74 cm

A = 0.9 + 0.74 = 1.64 cm

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