A skater weighing 50 kg, standing on the ice, throws a puck weighing 300 g in a horizontal direction at a speed of 15 m / sec.

| November 19, 2020 | education

A skater weighing 50 kg, standing on the ice, throws a puck weighing 300 g in a horizontal direction at a speed of 15 m / sec. How far will the skater roll if the coefficient of friction of the skates on the ice is 0.1?

mk = 50 kg.
mw = 300 g = 0.3 kg.
Vsh = 15 m / s.
μ = 0.1.
g = 9.8 m / s ^ 2.
S -?
Let’s find the speed of the skater after throwing the puck according to the law of conservation of momentum: msh * Vsh = mk * Vk.
Vk = msh * Vsh / mk.
Vk = 0.3 kg * 15 m / s / 50 kg = 0.09 m / s.
The skater moves only under the influence of friction force: mk * a = Ftr.
Ftr = μ * mk * g.
mk * a = μ * mk * g.
a = μ * g.
The skater’s movement will be determined by the formula: S = Vk ^ 2/2 * a.
S = Vk ^ 2/2 * μ * g.
S = (0.09 m / s) ^ 2/2 * 0.1 * 9.8 m / s ^ 2 = 0.004 m.
Answer: when throwing the puck, the skater will roll back a distance of S = 0.004 m.



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