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7. A block of mass 1.6 kg is moving across a smooth floor at 13.8 m/s and encounters a second block (initially at rest) of ma

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Answer #1

initial speed of the first block u1 = 13.8 m/s

mass m1 = 1.6 kg

mass of second block m2 = 3.4 kg

initial momentum m1u1 = 1.6*13.8 = 22.08

v1 and v2 , velocities of the two blocks after collision

Total momentum after collision = 1.6v1 + 3.4 v2 = 22.08 - conserve momentum.

elastic collision - Total mechanical energy is conserved

1/2*1.6 v12 + 1/2 *3.4 vs2 = 1/2 *1.6 *13.82  

1.6v12 + 3.4v22 = 304.7 ,

solving the two eq. we get

v1 = -4.97 m/s

v2 = 8.83 m/s

KE of the second block after collision = 1/2 *3.4*8.832 = 132.55 J

The block will move to the right and compress the spring and comes to rest instantaneously. At this instant the spring is compressed max and its potential energy is equal to the KE of the block.

1/2 kx2 = 1/m2v22  

x = (3.4/1250)1/2 * v2 = 0.46 m

amplitude of the oscillation A = 1.46 m - spring compression.

\omega = sqrt( k/m) =19.17 rads/s

Period T = 2\pi/\omega = 0.328 s.

time from equilibrium to max. displacement = T/4 = 0.082s

SHM : x(t) = 0.46Cos(19.17t +\pi/2) ,we take t=0 the instant of collision and the displacement x(0) =0

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