A spring-mass system is shown in the figure above. Mass B (300g) is initially 23

Question

A spring-mass system is shown in the figure above. Mass B (300g) is initially 23 cm from mass A (100g), traveling at 2.0 m s . The coefficient of kinetic friction with the surface is 0.21 and the coefficient of static friction is 0.54. Suppose the two blocks collide elastically and the spring constant is 32 N m

1. Calculate the speed of block B immediately before colliding with block A.

2. Suppose the collision takes a negligible amount of time so that friction may be ignored for the time of the collision. Calculate the speed of the two blocks immediately after the collision.

3. Calculate the maximum amount the spring is compressed.

4. Do the masses remain stopped after the spring is compressed the maximum amount? How do you know? Provide calculations to support your answer and reasoning.

Explanation / Answer

1] by work energy theorem, -umg x = o.5mv^2 - o.5mv1^2

v = sqrt(v1^2 - 2ugx)

= sqrt(2^2-2*0.21*9.8*0.23)

= 1.747 m/s

2] speed of blockB = [mB-mA]v1/[mA+mB] = 1.747*(300-100)/(300+100) = 0.8737 m/s

speed of blockA vA= [2mB]v1/[mA+mB] = 1.747*(2*300)/(300+100) = 2.6205 m/s

3] Let it be x, -umAgx - 0.5kx^2 = -0.5 mAvA^2

- 0.21*0.100*9.8*x - 0.5*32*x^2 = -0.5*0.100*2.6205^2

x = 0.1402 m answer

4] Spring force = kx = 32*0.1402 = 4.4864 N

maximum static friction force = us * mAg = 0.54*0.100*9.8 = 0.5292 N

Spring force is more, so mass will continue to move.

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