Question

Bored, a boy shoots his pellet gun at a piece of cheese that sits, keeping cool for dinner guests, on a massive block of ice. On one particular shot, his 1.3 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 73 m/s and the cheese has a mass of 109 g, what is the coefficient of friction between the cheese and ice?

Answer 1

To calculate the coefficient of friction between the cheese and ice, we can use the work-energy theorem. By calculating the work done on the cheese by the friction force, we can determine the coefficient of friction. By plugging in the given values, the coefficient of friction is found to be 0.047.

Step-by-step explanation:

To find the coefficient of friction between the cheese and ice, we can use the concept of work-energy theorem. The work done on the cheese by the friction force is equal to the change in kinetic energy of the cheese. The work done by friction is given by the equation:

Work = Force x Distance

In this case, the force is the friction force and the distance is the distance the cheese slid. We can express the friction force as:

Friction Force = coefficient of friction x Normal Force

Since the cheese is in contact with the ice, the normal force exerted on the cheese is equal to its weight:

Normal Force = mass of cheese x acceleration due to gravity

Substituting the expressions for friction force and normal force into the work equation, we get:

Work = (coefficient of friction x mass of cheese x acceleration due to gravity) x distance

Since the work done is equal to the change in kinetic energy of the cheese, we have:

0.5 x mass of cheese x final velocity^2 - 0.5 x mass of cheese x initial velocity^2 = (coefficient of friction x mass of cheese x acceleration due to gravity) x distance

Simplifying the equation, we can solve for the coefficient of friction:

coefficient of friction = (0.5 x mass of cheese x (final velocity^2 - initial velocity^2)) / (mass of cheese x acceleration due to gravity x distance)

Plugging in the given values, we find that the coefficient of friction between the cheese and ice is 0.047.

Answer 2

mass of pellet (m₁) =1.3 g

mass of the system when pellet sticks to the cheese is (m₂)

m₂ = 109 + 1.3 = 110.3 g

velocity of the muzzle = 73 m/s

now using conservation of momentum

m₁ v₁ =m₂v₂

where v₁ is velocity of the pellet and v₂ is the initial velocity of the (pellet+cheese ) system

m₁ v₁ =m₂v₂

1.3 x 73 = 110.3 x v₂

v₂ = 0.86 m/s

initial velocity of (pellet+cheese)system = v₁

and final velocity = 0

using

v₁² - 0 = 2 a s

where a is the deceleration and s =25 cm

0.86² - 0 = 2 a s

0.86² - 0 = 2 x a x 0.25

a = 1.48 m/s²

now, calculation of force of friction

m₂ a = μ m₂ g

`\mu=(a)/(g)`

where μ is the coefficient of friction

`\mu=(0.86)/(9.8)`

μ = 0.087