In Fig. 15-40, block 2 of mass 2.0 kg oscillates on the end of a spring in SHM with a period of 20 ms. The position of the block is given by x = (1.0 cm) cos (ωt + π/2). Block I of mass 4.0 kg slides toward block 2 with a velocity of magnitude 6.0 m/s, directed along the spring"s length. The two blocks undergo a completely inelastic collision at time t = 5.0 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?

Fig. 15-40

Question

Alexis Phelps

Physics

17 September, 22:32

In Fig. 15-40, block 2 of mass 2.0 kg oscillates on the end of a spring in SHM with a period of 20 ms. The position of the block is given by x = (1.0 cm) cos (ωt + π/2). Block I of mass 4.0 kg slides toward block 2 with a velocity of magnitude 6.0 m/s, directed along the spring"s length. The two blocks undergo a completely inelastic collision at time t = 5.0 ms. (The duration of the collision is much less than the period of motion.) What is the amplitude of the SHM after the collision?

Fig. 15-40

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Answers (1)

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Jesse Jordan · Answer 1

The correct answer for this question is this one:

First, we need to find k from

w^2 = k/m

so, we have

k = mw^2

k = 120852.3 N/m.

Next, solve for x: PE

PE = 0.5kx^2

x = sqrt (2PE/k).

So, from that, x is the answer.