Question

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 mass 3.4 kg in a fully elastic collision. The second block is attached to a spring of k = 1250 N/m. Assume the spring to be massless and does not interfere with the collision. After the collision, the second block is under simple harmonic motion. Determine, a. The amplitude of oscillation b. The frequency of oscillation. c. The time it takes to go from equilibrium to the maximum displacement. d. Functional form of SHM (i.e. x(t) = Acos(wt + p)

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 v_s² = 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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