In the figure, block 2 of mass 2.20 kg oscillates on the end of a spring in SHM

Question

In the figure, block 2 of mass 2.20 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 = (0.700 cm) cos(wt /2), Block 1 of mass 4.40 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 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

Explanation / Answer

in shm

v = A*w*sin(wt+pi/2)

w = 2pi/T = 2pi/0.012 = pi/0.006 rad/s

k = m2*w^2 = 2.2*(pi/0.006)^2 = 6.03*10^5 N/m

at t= 3 ms = 0.003 s

speed of block2 = u2 = -0.007*523.6*sin((pi0.003/0.006)+(pi/2))

u2 = 0

u1 = 9 m/s

m1 = 4.4 kg

momentum before collision = momentum after collision

m1*u1 + m2*u2 = ( m1 + m2)v

4.4*9 = (4.4+2.2)*V

v = 6 m/s

KE after collsion = max PE

0.5*(m1+m2)*v^2 = 0.5*k*A^2

0.5*(4.4+2.2)*6^2 = 0.5*6.03*10^5*A^2

A = 0.198 m = 1.98 cm

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