Problem P. 5078. (December 2018)
P. 5078. In an ice-hockey match it is not rare that the speed of the hockey puck reaches the value of 160 km/h.
\(\displaystyle a)\) How far could a puck having the above stated speed move on the ice if the coefficient of kinetic friction between the puck and the ice is 0.1?
\(\displaystyle b)\) By what average force must the puck be shot in order to accelerate it to this speed? The stick and the puck are in contact for approximately 0.01 s. The weight of the puck is approximately 1.5 N.
One of the skaters of the local team – preparing for the penalty shot – starts to move with the puck. He decides that he will shoot the puck from a distance of 5 m from the goal line. He knows that the goal-tender of the opposing team has an excellent reaction time of 0.15 s.
\(\displaystyle c)\) By what initial speed should he shoot the puck in order that it reaches the net before the goal-tender starts to move?
\(\displaystyle d)\) By what force should he shoot the puck?
(4 pont)
Deadline expired on January 10, 2019.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. \(\displaystyle a)\) A korong kezdősebessége kb. \(\displaystyle v_0=44\) m/s, lassulása \(\displaystyle a=-\mu g\approx -1~\rm m/s^2\). Az egyenletesen lassuló korong a megállásáig \(\displaystyle s=\frac{v_0^2}{2\vert a\vert}\approx 1000~\rm m\) utat tenne meg.
\(\displaystyle b)\) A korong tömege: \(\displaystyle m=0{,}15~\)kg. A \(\displaystyle \Delta t=0{,}01\) másodpercig ható erő akkor gyorsítja fel \(\displaystyle v_0\) sebességre, ha az átlagos erő nagysága
\(\displaystyle F=\frac{mv_0}{\Delta t}\approx 670~\rm N.\)
\(\displaystyle c)\) A korong 5 méteres elmozdulása sokkal kisebb, mint az \(\displaystyle a)\) kérdésben szereplő 1000 méter, emiatt (ezen a távon) a korong lassulása figyelmen kívül hagyható. A korong sebessége legalább
\(\displaystyle v_1=\frac{5~\rm m}{0{,}15~\rm s}=33~\frac{\rm m}{\rm s}= 120~\frac{\rm km}{\rm h}.\)
\(\displaystyle d)\) Mivel \(\displaystyle v_1/v_0=0{,}75,\) a kifejtendő átlagerő \(\displaystyle 0{,}75\cdot 670~\rm N\approx 500~{\rm N}.\)
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Problems in Physics of KöMaL, December 2018