Two equal mass hockey pucks collide in a perfectly elastic collision. The first
puck is stationary, and the second puck collides with the first while
travelling 15 m/s [E]. The first puck travels in the direction of [E4°S], while the second puck travels with an unknown velocity, but with a direction of [E18°N]. Determine the final velocities of both pucks.
conserve momentum. Since the masses are equal, just work with velocities.
0 + 15<1,0> = v1<cos4°,-sin4°> + v2<cos18°,sin18°>
now solve for v1 and v2
To solve this problem, we will use the principles of conservation of momentum and conservation of kinetic energy.
Let's define the variables:
- m: mass of each hockey puck
- v1i: initial velocity of the first puck (which is stationary)
- v2i: initial velocity of the second puck (15 m/s [E])
- θ2: angle of the initial velocity of the second puck ([E18°N])
Step 1: Calculate the x and y components of the initial velocities.
Given the angle θ2 = [E18°N], we can decompose the initial velocity v2i into its x and y components.
V2ix = v2i * cos(θ2)
V2iy = v2i * sin(θ2)
Step 2: Apply conservation of momentum in the x direction.
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the first puck is stationary (v1i = 0), the momentum in the x direction is preserved.
m * v1f + m * V2ix = m * v1f' + m * V2fx
Step 3: Apply conservation of momentum in the y direction.
The momentum in the y direction is also preserved.
0 + m * V2iy = m * v1f' * sin(θ1) + m * V2fy
Step 4: Apply 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.
(1/2) * m * v1f^2 + (1/2) * m * v2i^2 = (1/2) * m * v1f'^2 + (1/2) * m * V2f^2
Step 5: Solve the system of equations.
Let's solve the system of equations to find the final velocities v1f and V2f.
From step 2:
m * v1f = - m * V2ix (equation 1)
From step 3:
m * V2iy = m * v1f' * sin(θ1) + m * V2fy (equation 2)
From step 4:
(1/2) * m * v1f^2 + (1/2) * m * v2i^2 = (1/2) * m * v1f'^2 + (1/2) * m * V2f^2
(v1f^2 + v2i^2) = (v1f'^2 + V2f^2) (equation 3)
Since the pucks have equal masses, we can simplify the equations:
From equation 1:
v1f = - V2ix (equation 4)
From equation 2:
V2iy = v1f' * sin(θ1) + V2fy (equation 5)
From equation 3:
v1f^2 + v2i^2 = v1f'^2 + V2f^2 (equation 6)
Step 6: Substitute the values and solve for the unknowns.
Substitute the given values and unknowns into the equations and solve for the final velocities.
From equation 4, substituting V2ix = 15 m/s * cos(18°):
v1f = - (15 m/s * cos(18°))
From equation 5:
V2iy = v1f' * sin(θ1) + V2fy
Since we don't have the angle θ1 or the y component of the initial velocity V2fy, we cannot solve for V2iy. However, we can solve for v1f' using equation 6:
(v1f^2 + v2i^2) = (v1f'^2 + V2f^2)
[(-15 m/s * cos(18°))^2 + (15 m/s)^2] = (v1f'^2 + V2f^2)
Solve for v1f', which is the final velocity of the first puck (in m/s).
Once we have v1f', we can calculate V2f using equation 5:
V2f = V2iy - v1f' * sin(θ1)
By substituting the known values and solving these equations, we can find the final velocities of both pucks.
To solve this problem, we need to apply the principles of conservation of momentum and conservation of kinetic energy.
First, let's break down the initial velocities of both pucks into their x and y components.
For the first puck (initially stationary):
- Initial velocity in the x direction (Vx1_initial) = 0 m/s
- Initial velocity in the y direction (Vy1_initial) = 0 m/s
For the second puck:
- Initial velocity (V2_initial) = 15 m/s [E]
- The direction of [E18°N] means the angle between the velocity vector and the positive x-axis is 18 degrees.
We can find the x and y components of the second puck's initial velocity using trigonometry, assuming that the positive x-axis is the reference axis:
- V2_initial_x = V2_initial * cos(theta)
- V2_initial_y = V2_initial * sin(theta)
V2_initial_x = 15 m/s * cos(18°) = 14.3 m/s [E]
V2_initial_y = 15 m/s * sin(18°) = 4.2 m/s [N]
Now, let's denote the final velocities of the first and second puck as V1_final and V2_final, respectively.
Since it is a perfectly elastic collision, momentum and kinetic energy are conserved.
Using the conservation of momentum:
- Initial momentum = Final momentum
The initial momentum is given by:
- Initial momentum of the first puck = m1*v1_initial (since the first puck is stationary, the initial momentum is 0)
- Initial momentum of the second puck = m2*v2_initial_x
The final momentum is given by:
- Final momentum of the first puck = m1*v1_final
- Final momentum of the second puck = m2*v2_final_x
Since the masses of the pucks are equal (m1 = m2 = m), we can rewrite the conservation of momentum equation as:
m2 * v2_initial_x = m1 * v1_final + m2 * v2_final_x
Now, let's use the conservation of kinetic energy:
- Initial kinetic energy = Final kinetic energy
The initial kinetic energy is given by:
- Initial kinetic energy of the first puck = (1/2) * m1 * (v1_initial)^2
- Initial kinetic energy of the second puck = (1/2) * m2 * (v2_initial_x)^2 + (1/2) * m2 * (v2_initial_y)^2
The final kinetic energy is given by:
- Final kinetic energy of the first puck = (1/2) * m1 * (v1_final)^2
- Final kinetic energy of the second puck = (1/2) * m2 * (v2_final_x)^2 + (1/2) * m2 * (v2_final_y)^2
Using the conservation of kinetic energy equation, we can write:
(1/2) * m2 * (v2_initial_x)^2 + (1/2) * m2 * (v2_initial_y)^2 = (1/2) * m1 * (v1_final)^2 + (1/2) * m2 * (v2_final_x)^2 + (1/2) * m2 * (v2_final_y)^2
Now, we have two equations:
1) m2 * v2_initial_x = m1 * v1_final + m2 * v2_final_x
2) (1/2) * m2 * (v2_initial_x)^2 + (1/2) * m2 * (v2_initial_y)^2 = (1/2) * m1 * (v1_final)^2 + (1/2) * m2 * (v2_final_x)^2 + (1/2) * m2 * (v2_final_y)^2
Solving these equations simultaneously will give us the final velocities of both pucks: V1_final and V2_final.