Curling Sport Momentum and Collisions Physics Problem

Question

Disappointed with a 4th place tin medal in bobsledding, Burl has decided to take up curling. One of the opposing team's stones is resting on the outer edge of the house as shown below. With which speed should Burl hit the other team's rock in order to knock it out of the way and land in the middle of the target?

The target has a radius(R) of 3.65m (the rock happens to be lying such that x and y are equal distances) and the stones each have a mass(m) of 17kg. The coefficient of friction between the ice and granite stones is 0.012. Assume that Burl is able to hit the stone at the correct angle to put his rock on a straight line course to the bullseye and that his incoming stone will slide completely in the y-direction as shown.Also, assume an elastic collision.

y

|

|______x

Target=O

X

O-------

\ | Y

R \ |

o opponent's stone

o burl's stone

Problem looks like this ibb.co/6nHQMfm

Answer 1

Let

v₀ (to be calculated) be the velocity with which Burl's stone should hit the other,

v (unknown) be the velocity of Burl's stone after collision,

w (unknown) be the velocity of the other stone after collision,

R = 3.65 be the distance to travel from the collision to the target,

m = 17 kg be the mass of the stones,

f = 0.012 be the coefficient of friction,

g = 9.81 m/s² be gravitational acceleration.

This is the general approach:

We calculate v using conservation of energy --

the stones kinetic energy right after the collision must be completely dissipated by friction.

From v we then calculate v₀ by conservation of momentum.

CALCULATE v:

The stone's weight is mg.

The force of friction is mgf.

The work done by friction in traversing distance R is Rmgf.

(This work is also the stone's energy dissipated by friction.)

The kinetic energy of the stone right after the collision is mv²/2.

The kinetic energy of the stone when it reaches the center is 0.

By conservation of energy

mv²/2 = Rmgf

v = √(2Rgf)

CALCULATE v₀:

Let v_x be the x-component of the vector v,

Let v_y be the y-component of the vector v,

Let w_x be the x-component of the vector w

Let w_y be the y-component of the vector w.

The x-component of v₀ equals 0, and the y-component is v₀.

The above notation means in particular that, by Pythagorean theorem,

v² = v_x² + v_y² (0)

w² = w_x² + w_y²

Before the collision, only Burl's stone contributes to the momentum of the system.

The momentum in the x-direction is 0, and

the momentum in the y-direction mv₀.

After the collision both stones contribute to the system's momentum,

and their total momentum must equal the momentum before the collision.

The equality of momenta in the x-direction is expressed by the equation

mv_x + mw_x = 0

which simplifies into

v_x + w_x = 0 (1)

The equality of momenta in the y-direction is expressed by the equation

mv_y + mw_y = mv₀

which simplifies into

v_y + w_y = v₀ (2)

Below we will use the square of this equation

v_y² + 2v_yw_y + w_y² = v₀² (3)

The information that the collision is elastic is defined to mean that energy is conserved.

That means

mv²/2 + mw²/2 = mv₀²/2

which simplifies into

v² + w² = v₀²

Expressed in terms of the x and y components

v_x² + v_y² + w_x² + w_y² = v₀²

Substituting w_x = -v_x from equation (1)

v_x² + v_y² + v_x² + w_y² = v₀²

Eliminating v₀ using equation (3)

v_x² + v_y² + v_x² + w_y² = v_y² + 2v_yw_y + w_y²

Which simplifies into

v_x² = v_yw_y (4)

The above allows many solutions.

I suspect that your teacher wants the solution where v_x = v_y,

although that is in contradiction with the given figure.

In the given figure, the other ball rests in a spot with equal x and y coordinates.

Burl's ball will strike it somewhere to the left and below,

making it impossible for v_x = v_y.

There are two ways to justify the assumption v_x = v_y.

The first way is to declare the figure in error:

the other ball should rest to the side, so that at the time of collision Burl's ball's

center has equal x and y coordinates.

The second way is to declare our result only an approximation;

the smaller are the balls, the more accurate will be the approximation.

From now on assume that v_x = v_y.

Then equation (4) simplifies into

v_x = w_y

Using it in equation (2)

v₀ = 2v_x

And using equation (0)

v₀ = v√2

Combine that with the result for v

v₀ = √(2Rgf) √2 = 2√(Rgf)

Substituting actual numbers

v₀ = 2√(3.65×9.81×0.012) = 1.31 m/s

Notice that the answer is independent of the mass m, which is given,

but should depend on the size of the balls, which is not given.