A mass \(M=0.460 \mathrm{~kg}\) moves with an initial speed \(v=3.20 \mathrm{~m} / \mathrm{s}\) on a level frictionless air track. The mass is initially a distance \(D=0.250 \mathrm{~m}\) away from a spring with \(k=\) \(840 \mathrm{~N} / \mathrm{m}\), which is mounted rigidly at one end of the air track. The mass compresses the spring a maximum distance \(d\), before reversing direction. After bouncing off the spring the mass travels with the same speed \(v\), but in the opposite dircction. a) Determine the maximum distance that the spring is compressed. b) Find the total elapsed time until the mass returns to its starting point. (Hint: The mass undergoes a partial cycle of simple harmonic motion while in contact with the spring.)

Step by step solution

01

Apply the conservation of energy principle

The initial kinetic energy of the mass is equal to the potential energy stored in the spring at maximum compression. This can be written as: $$ \frac{1}{2}Mv^2 = \frac{1}{2}kd^2 $$

02

Solve for d

We can solve the equation above for d: $$ d=\sqrt{\frac{Mv^2}{k}} $$

03

Calculate d

Now we can plug in the values given in the problem: $$ d = \sqrt{\frac{(0.460\,\text{kg})(3.20\,\text{m/s})^2}{840\,\text{N/m}}} $$ Calculate the numerical value of d: $$ d \approx 0.0930\,\text{m} $$ #b) Find the total elapsed time until the mass returns to its starting point.#

04

Calculate the time spent compressing the spring

When the mass is compressing the spring, it undergoes simple harmonic motion. The period of simple harmonic motion can be calculated using the formula: $$ T = 2\pi\sqrt{\frac{M}{k}} $$ The mass undergoes 1/2 of a full cycle while compressing the spring, so the time spent compressing is 1/2 the period: $$ t_\text{compress}=\frac{1}{2} \cdot 2\pi\sqrt{\frac{M}{k}} = \pi\sqrt{\frac{M}{k}} $$

05

Calculate the time spent moving back to the starting point

Once the mass bounces back, it travels back to the starting point with the same speed as it initially had. We can calculate the time it takes for the mass to travel back using the formula: $$ t_\text{return} = \frac{D}{v} $$

06

Calculate the total elapsed time

Now we can calculate the total elapsed time by adding the time spent compressing the spring and the time spent moving back to the starting point: $$ t_\text{total} = t_\text{compress}+t_\text{return} $$ Plugging in the values we found: $$ t_\text{total} = \pi\sqrt{\frac{(0.460\,\text{kg})}{(840\,\text{N/m})}} + \frac{(0.250\,\text{m})}{(3.20\,\text{m/s})} $$ Calculate the numerical value of the total elapsed time: $$ t_\text{total} \approx 0.230\,\text{s} $$

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