a 80.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 47.0 m/s. if both are initially at rest and if the ice is frictionless, how far (in m) does the player recoil in the time it takes the puck to reach the goal 18.0 m away?



Answer :

ayune

As a result of the law of conservation of momentum, the man will recoil as far as 0.034 m.

The law of conservation of momentum:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Where:

u₁ , u₂ = initial velocity of object 1 and object 2

v₁ , v₂ = final velocity of object 1 and object 2

Since both the player and the puck are initially at rest, hence  u₁ = 0, u₂ = 0.

Therefore,

m₁v₁ + m₂v₂ = 0

Parameters given:

m₁ = 80 kg

m₂ = 0.150 kg

v₂ = 47 m/s

Plug the parameters into the equation,

80 . v₁ + 0.15 . 47 = 0

v₁ = - 0.088 m/s

Time the puck reach 18 m goal:

t = distance / velocity = 18 / 47 = 0.38 s

The player will recoil:

distance = velocity x t = 0.088 x  0.38 = 0.034 m = 3.4 cm

Learn more about momentum here:

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