Question 19.9: Skiing and vibrations Imagine that you ski down a slope wear......
Skiing and vibrations
Imagine that you ski down a slope wearing a Velcro ski vest and then continue skiing on a horizontal surface at the bottom of the hill. There you run into a padded, Velcro covered cart, which is also on skis (see the figure below). A 1280 N/m spring is attached to the other end of the cart and also to a wall. The spring compresses after your 60-kg body hits and sticks to the 20-kg cart. Your speed is 16 m/s just before you hit the cart. (a) What is your maximum speed after joining with the cart? (b) What is the maximum compression of the spring? (c) What is the period of the vibrational motion? (d) What is your maximum acceleration and where does it occur?
Sketch and translate Draw a labeled sketch (see the first figure at right). (a) The collision with the cart is a completely inelastic collision because you stick to the cart. If you and the cart are the system, then during the moment of the collision, the momentum of the system is constant (the same before and after the collision). We can use impulse-momentum ideas to determine your speed immediately after the collision. (b) Now choose a different system: you, the cart, and the spring. The kinetic energy of you and the cart is converted into the spring’s elastic potential energy as the spring compresses. (c) This is now a spring-object system that vibrates— you can determine the period of the vibration and the frequency. (d) The maximum acceleration is when the spring is compressed or stretched the most—when x=\pm A . We can determine the magnitude of the maximum acceleration using Newton’s second law.
Simplify and diagram Assume that the snowy surface is perfectly smooth (zero friction), and that the spring has zero mass and obeys Hooke’s law. (a) A momentum bar chart your collision with the cart (see the first figure at the right). There are no horizontal forces exerted on the you+ cart system along the x-direction during the instant of collision. Thus the x component of momentum is constant. (b) Immediately after the collision, you and the cart, now joined, compress the spring. An energy bar chart represents this process (see the second figure above).
Represent mathematically (a) Use momentum constancy for the collision with the cart to find the speed of you+cart after the collision:
m_{\text {You }} v_{\text {You i } x}+m_{\text {Cart }} v_{\text {Cart } i \text{ }x}=m_{\text {You }+\text { Cart }} v_{\mathrm{f} \text{ }x}
(b) Use energy constancy to determine the maximum compression distance:
\frac{1}{2} m_{\text {You }+\text { Cart }} v_{\mathrm{f} x}^2=\frac{1}{2} k A^2
(c) Determine the period using
T=2 \pi \sqrt{\frac{m_{\mathrm{You}+\mathrm{Cart}}}{k}}
(d) Use Newton’s second law to determine the maximum acceleration during the vibrations after the collision:
a_x=\frac{-k x}{m_{\mathrm{You}+\mathrm{Cart}}}
a is maximum when the spring has its maximum compression (when x = -A).
Solve and evaluate (a) The velocity of you and the cart immediately after the collision is
v_{\mathrm{f} x}=\frac{m_{\text {You }} v_{\text {You } i \text{ }x}+m_{\text {Cart }} v_{\text {Cart } i\text{ } x}}{m_{\text {You }+\text { Cart }}}=\frac{(60 \mathrm{~kg})(16 \mathrm{~m} / \mathrm{s})+0}{(60 \mathrm{~kg}+20 \mathrm{~kg})}
= 12 m/s
The speed seems reasonable, and the sign matches the original direction of motion.
(b) Using the equation \frac{1}{2} m_{\text {You }+\text { Cart }} v_{\mathrm{f}}^2=\frac{1}{2} k A^2 , we can cancel the (1/2), divide both sides of the equation by k, and take the square root. We find that the maximum displacement of the cart from its equilibrium position is
A=\sqrt{\frac{m_{\text {You }+\mathrm{Cart}}}{k}} v_{\mathrm{f}}=\sqrt{\frac{80 \mathrm{~kg}}{1280 \mathrm{~N} / \mathrm{m}}}(12 \mathrm{~m} / \mathrm{s})=3.0 \mathrm{~m}
The compression seems a little high but the skier was moving with high speed; the units are fine.
(c) The period for the you+cart vibration is
T=2 \pi \sqrt{\frac{m_{\text {You }+\text { Cart }}}{k}}=2 \pi \sqrt{\frac{80 \mathrm{~kg}}{1280 \mathrm{~N} / \mathrm{m}}}=1.6 \mathrm{~s}
(d) The cart started at its equilibrium position 1x = 02 and was moving at 12 m/s after you collided with it. This is its maximum speed. The amplitude of the vibration was 3.0 m. Thus, the maximum acceleration of the you+ cart system when at the x = -A position is
a_{\max }=\frac{-k(-A)}{\left(m_{\text {You }+\text { Cart }}\right)}=\frac{(1280 \mathrm{~N} / \mathrm{m})(3.0 \mathrm{~m})}{(60 \mathrm{~kg}+20 \mathrm{~kg})}=48 \mathrm{~m} / \mathrm{s}^2
The maximum acceleration of you and the cart is five times free-fall acceleration. The units for the result are the units of acceleration: \frac{(\mathrm{N} / \mathrm{m}) \times(\mathrm{m})}{(\mathrm{kg})}=\frac{\mathrm{N}}{\mathrm{kg}}=\frac{\mathrm{kg} \times \mathrm{m} / \mathrm{s}^2}{\mathrm{~kg}}=\mathrm{m} / \mathrm{s}^2 .
Try it yourself: (a) How would the period of vibration change if you hit the cushion at half the velocity in the example? (b) What would be the period of vibration for a person whose mass is half of yours and whose speed before the collision with the cart is 0.8 times your speed?
Answer: (a) The period is independent of the amplitude. (b) The period will be 1.24 s.