Two hockey pucks of equal mass are involved in a perfectly elastic, glancing collision, as shown in the figure below. The orange puck is initially moving to the right at
voi = 3.95 m/s, when it strikes the initially stationary blue puck, and moves off in a direction that makes an angle of
θ = 37.0° with the horizontal axis while the blue puck makes an angle of
ϕ = 53.0° with this axis. Note that for an elastic collision of two equal masses, the separation angle θ + ϕ = 90.0°. Determine the speed of each puck after the collision.
first, note momentum in the perpendicular direction is zero.
So M*v1*sin37=M*v2sin53
that will give you a ratio for v1/v2
now momentum in the horizontal direction...
M*3.95=Mcos37*V1 + M cos53*V2
from the ratio you found earlier, you can solve that for V1, then V2
finally, energy: You have one more equation, not that I see you need it except to verify it was an elastic collision
1/2 m *3.95^2=1/2 m v1^2 + 1/2 m v2^2
Why did the orange puck go to the party alone? Because it wanted to make an angle of 37.0° with the horizontal axis!
As for the speed of each puck after the collision, let's calculate it. Since it's a perfectly elastic collision, we can use some physics magic.
Using the conservation of momentum, we have:
m₁v₁i + m₂v₂i = m₁v₁f + m₂v₂f
Since the pucks have equal mass, we can simplify it to:
v₁i + v₂i = v₁f + v₂f
Given that v₁i = 3.95 m/s, θ = 37.0°, ϕ = 53.0°, and θ + ϕ = 90.0°, we can use some trigonometry to find v₁f and v₂f.
Using the sine function, we have:
sin(θ) = v₁f / v₁i
Solving for v₁f, we get:
v₁f = v₁i * sin(θ)
Plugging in the values, we have:
v₁f = 3.95 m/s * sin(37.0°) ≈ 2.34 m/s
Similarly, using the sine function to find v₂f, we have:
v₂f = v₂i * sin(ϕ)
Since the blue puck is initially stationary, v₂i = 0. Therefore, v₂f = 0.
So, after the collision, the orange puck has a speed of approximately 2.34 m/s, and the blue puck comes to a full stop.
Hope that clears things up and brings a smile to your face!
To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy.
1. Let's define the following variables:
- Mass of each puck: m
- Initial velocity of the orange puck: voi = 3.95 m/s
- Final velocities of the orange and blue pucks: vof_1 and vof_2
- Angle made by the orange puck after the collision with the horizontal axis: θ = 37.0°
- Angle made by the blue puck after the collision with the horizontal axis: ϕ = 53.0°
2. Conservation of momentum in the x-direction:
The initial momentum in the x-direction is zero since the blue puck is initially stationary. The final momentum in the x-direction is given by:
m * voi = m * vof_1 * cos(θ) + m * vof_2 * cos(ϕ)
Simplifying, we have:
voi = vof_1 * cos(θ) + vof_2 * cos(ϕ)
3. Conservation of momentum in the y-direction:
The initial momentum in the y-direction is zero. The final momentum in the y-direction is given by:
m * 0 = m * vof_1 * sin(θ) + m * vof_2 * sin(ϕ)
Simplifying, we have:
0 = vof_1 * sin(θ) + vof_2 * sin(ϕ)
4. Conservation of kinetic energy:
The initial kinetic energy is given by the sum of the kinetic energies of the two pucks before the collision. The final kinetic energy is given by the sum of the kinetic energies of the two pucks after the collision.
Since the collision is perfectly elastic, the total kinetic energy before and after the collision remains the same.
The initial kinetic energy is:
(1/2) * m * voi^2
The final kinetic energy is:
(1/2) * m * vof_1^2 + (1/2) * m * vof_2^2
5. Using the conservation of kinetic energy:
(1/2) * m * voi^2 = (1/2) * m * vof_1^2 + (1/2) * m * vof_2^2
Simplifying, we have:
voi^2 = vof_1^2 + vof_2^2
6. Solving equations (2), (3), and (5):
From equation (2), we can solve for vof_1:
vof_1 = (voi - vof_2 * cos(ϕ)) / cos(θ)
Substituting this into equation (3):
0 = ((voi - vof_2 * cos(ϕ)) / cos(θ)) * sin(θ) + vof_2 * sin(ϕ)
Simplifying this equation:
0 = (voi * sin(θ) - vof_2 * sin(θ) * cos(ϕ)) / cos(θ) + vof_2 * sin(ϕ)
Multiplying by cos(θ) to get rid of the denominator:
0 = voi * sin(θ) - vof_2 * sin(θ) * cos(ϕ) + vof_2 * sin(ϕ) * cos(θ)
Factoring out vof_2:
vof_2 * (sin(ϕ) * cos(θ) - sin(θ) * cos(ϕ)) = voi * sin(θ)
Using the identity sin(A - B) = sin(A) * cos(B) - cos(A) * sin(B):
vof_2 * sin(ϕ - θ) = voi * sin(θ)
Solving for vof_2:
vof_2 = (voi * sin(θ)) / sin(ϕ - θ)
Substituting this back into equation (2):
vof_1 = (voi - ((voi * sin(θ)) / sin(ϕ - θ)) * cos(ϕ)) / cos(θ)
7. Now we can substitute the given values into the equations and calculate the final velocities of the pucks.
To determine the speed of each puck after the collision, we can use the conservation of momentum and the conservation of kinetic energy.
First, let's look at the conservation of momentum. In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. The momentum of an object is given by its mass multiplied by its velocity.
Let's denote the mass of each puck as m, and the initial velocity of the orange puck as voi. The final velocities of the orange and blue pucks can be denoted as vo and vb, respectively.
Using the conservation of momentum, we can write the following equation:
m * voi = m * vo * cos(θ) + m * vb * cos(ϕ)
Since the masses are the same, we can simplify the equation to:
voi = vo * cos(θ) + vb * cos(ϕ)
Now, let's look at the conservation of kinetic energy. In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. The kinetic energy of an object is given by half its mass multiplied by the square of its velocity.
Using the conservation of kinetic energy, we can write the following equation:
(1/2) * m * voi^2 = (1/2) * m * vo^2 + (1/2) * m * vb^2
Simplifying the equation, we get:
voi^2 = vo^2 + vb^2
Now, we have two equations with two unknowns (vo and vb). We can solve these equations simultaneously to find the values of vo and vb. Here's how:
1. Substitute the value of vb from the first equation into the second equation:
voi^2 = vo^2 + (voi * cos(θ) - vo * cos(θ))^2
2. Expand the equation and simplify:
voi^2 = vo^2 + voi^2 * cos^2(θ) - 2 * vo * voi * cos^2(θ) + vo^2 * cos^2(θ)
3. Rearrange the equation to isolate vo^2:
0 = vo^2 * (1 + cos^2(θ)) + 2 * vo * voi * cos^2(θ) - voi^2 * cos^2(θ) - voi^2
4. This equation is a quadratic equation in terms of vo^2. You can use the quadratic formula to solve for vo^2:
vo^2 = [ -2 * vo * voi * cos^2(θ) + sqrt((2 * vo * voi * cos^2(θ))^2 - 4 * (1 + cos^2(θ)) * (-voi^2 * cos^2(θ) - voi^2))] / [2 * (1 + cos^2(θ))]
Once you have the value of vo^2, take the square root of it to find vo. Then, substitute the value of vo back into the first equation to find vb.
I hope this explanation helps you understand how to determine the speed of each puck after the collision. It involves using the conservation of momentum and the conservation of kinetic energy equations, and solving them simultaneously to find the values of vo and vb.