Bored, a boy shoots his pellet gun at a piece of cheese that sits on a massive block of ice. On one particular shot, his 1.2-g pellet gets stuck in the cheese, causing it to slide 25 cm before coming to a stop. If the muzzle velocity of the gun is known to be 65 m>s, and the cheese has a mass of 120 g, what is the coefficient of friction between the cheese and the ice?

0.084

Explanation:

from the question we are given the following:

mass of pellet (M₁) = 1.2 g = 0.0012 kg

mass of cheese (Mc) = 120 g = 0.12 kg

total mass (M₂) = 0.1212 kg

distance (s) = 25 cm = 0.25 m

initial velocity (U₁) = 65 m/s

final velocity (U₂) = ?

coefficient of friction = ?

To get the coefficient of friction we have to apply the formula below:

frictional force = coefficient of friction x total mass x acceleration due to gravity

where

frictional force = mass x acceleration (a)

so our equation becomes

mass x acceleration (a) = coefficient of friction x mass x acceleration due to gravity

and finally we have

coefficient of friction = / frac{acceleration}{acceleration due to gravity}

so we need to find the acceleration of the pellet and cheese

from the conservation of momentum we can say

M₁ x U₁ = M₂ x U₂

0.0012 x 65 = 0.1212 x U₂

U₂ = 0.64 m/s

from the equation of motion

U₂^{2} = U₁^{2} + 2as

U₂ = final velocity = 0 since the pellet and cheese come to a stop

we now have

-U₁^{2} = 2as

(-0.64) ^{2} = 2 X a X 0.25

a = 0.82 m/s^{2}

now that we have a we can substitute it into coefficient of friction = / frac{acceleration}{acceleration due to gravity}

coefficient of friction = / frac{0.82}{9.8} = 0.084