An elastic collision occurs between two air hockey pucks in which one puck is at rest and the other is moving with a speed of 0.7 m/s. After the collision, the puck initially in motion makes an angle of 20.00 o with its original direction, and the struck puck moves at an angle of 70.00 o on the other side of the original direction. What is the final speed of the first puck? What is the speed of the second?
please solve this ..i don't want only theory explanation so please completly solve it
I will be happy to critique your solution, I am not about to do it and allow you to copy it.
To solve this problem, let's simplify the scenario by assuming the mass of both pucks is the same. This assumption allows us to use the conservation of momentum and conservation of kinetic energy equations to solve for the final velocities.
Let's denote the velocity of the initially moving puck as v₁, and the velocity of the initially stationary puck as v₂. The angles of deflection are given as 20.00° and 70.00°, respectively.
Step 1: Conservation of Momentum
In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision:
m₁ * v₁ = m₁ * v₁f * cos(20°) + m₂ * v₂f * cos(70°) (Equation 1)
Here, v₁f is the final velocity of the initially moving puck, and v₂f is the final velocity of the initially stationary puck.
Since we assume both pucks have the same mass, we can simplify Equation 1 to:
v₁ = v₁f * cos(20°) + v₂f * cos(70°) (Equation 2)
Step 2: 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) * m₁ * v₁² = (1/2) * m₁ * v₁f² + (1/2) * m₂ * v₂f² (Equation 3)
Again, since we assume both pucks have the same mass, we can simplify Equation 3 to:
v₁² = v₁f² + v₂f² (Equation 4)
Step 3: Solve for v₁f and v₂f
To solve the system of equations (Equations 2 and 4), we can substitute v₂f from Equation 2 into Equation 4:
v₁² = v₁f² + (v₁ - v₁f * cos(20°))² * cos(70°) (Equation 5)
Expanding Equation 5 and simplifying, we have:
0 = v₁f² - 2 * v₁f * v₁ * cos(20°) + v₁² - 2 * v₁f² * cos(20°) * cos(70°) + v₁² * cos²(70°)
By rearranging terms, we get the quadratic equation:
(1 - cos²(70°)) * v₁f² - 2 * v₁f * v₁ * cos(20°) + (1 + cos²(70°)) * v₁² = 0
So the quadratic equation is:
(0.342 - 0.342) * v₁f² - 1.851 * v₁f * v₁ + (0.342 + 0.658) * v₁² = 0
We can simplify this to:
0.658 * v₁² - 1.851 * v₁f * v₁ = 0
Now, we can solve this quadratic equation for v₁f:
v₁f = (1.851 * v₁) / 0.658 (Equation 6)
Step 4: Solve for v₁f and v₂f
Using Equation 6, we can substitute v₁f into Equation 2 to solve for v₂f:
v₁ = (1.851 * v₁) / 0.658 * cos(20°) + v₂f * cos(70°)
Simplifying the equation:
v₁ * 0.658 = 1.851 * v₁ * cos(20°) + 0.658 * v₂f * cos(70°)
0.658 * v₁ * (1 - cos(20°)) = 0.658 * v₂f * cos(70°)
v₂f = v₁ * (1 - cos(20°)) / cos(70°) (Equation 7)
Step 5: Plug in values and calculate
Now, let's substitute the given values into Equations 6 and 7 to calculate the final velocities:
v₁ = 0.7 m/s
angle1 = 20.00°
angle2 = 70.00°
Using Equation 6:
v₁f = (1.851 * 0.7) / 0.658 ≈ 1.972 m/s
Using Equation 7:
v₂f = 0.7 * (1 - cos(20°)) / cos(70°) ≈ 0.533 m/s
Therefore, after the collision, the final speed of the first puck (initially at rest) is approximately 1.972 m/s, and the final speed of the second puck (initially moving at 0.7 m/s) is approximately 0.533 m/s.