After a completely inelastic collision, two objects of the same mass and same initial speed are found to move away together at 1/5 their initial speed. Find the angle between the initial velocities of the objects.
To find the angle between the initial velocities of the objects after the collision, we can use the conservation of momentum and the conservation of kinetic energy.
In an inelastic collision, the objects stick together and move as one unit after the collision. This means that the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the initial velocities of the two objects as v₁ and v₂, and their final velocity together as v.
According to the conservation of momentum:
m₁v₁ + m₂v₂ = (m₁ + m₂)v
Since the two objects have the same mass (m₁ = m₂), this equation simplifies to:
v₁ + v₂ = 2v
Now, let's use the conservation of kinetic energy. In an inelastic collision, some kinetic energy is lost due to the deformation or other factors. The kinetic energy is given by the formula: K.E. = 0.5 * mass * velocity^2.
The initial kinetic energy before the collision is:
K.E.₁ = 0.5 * m₁ * v₁^2 + 0.5 * m₂ * v₂^2
The final kinetic energy after the collision is:
K.E.₂ = 0.5 * (m₁ + m₂) * v^2
Since the objects move away together at 1/5 their initial speed, the final kinetic energy can be expressed as:
K.E.₂ = 0.5 * (m₁ + m₂) * (1/5)^2 * (v₁^2 + v₂^2)
Setting K.E.₁ = K.E.₂ and substituting the values, we get:
0.5 * m₁ * v₁^2 + 0.5 * m₂ * v₂^2 = 0.5 * (m₁ + m₂) * (1/5)^2 * (v₁^2 + v₂^2)
Simplifying this equation, we have:
v₁^2 + v₂^2 = (1/5)^2 * (v₁^2 + v₂^2)
Solving for v₁^2 + v₂^2, we get:
24 * v₁^2 + 24 * v₂^2 = 5 * v₁^2 + 5 * v₂^2
Now, we can substitute the value of v₁ + v₂ from the conservation of momentum into this equation.
v₁^2 + v₂^2 = 5/24 * (v₁ + v₂)^2
Substituting v₁ + v₂ = 2v, we get:
v₁^2 + v₂^2 = 5/24 * (2v)^2
Simplifying further, we have:
v₁^2 + v₂^2 = 5/24 * 4v^2
v₁^2 + v₂^2 = 5/6 * v^2
Now, we can divide both sides of this equation by v^2 to get rid of the v term:
(v₁^2 + v₂^2) / v^2 = 5/6
Finally, we can express this equation in terms of the angle between the initial velocities (θ) using the trigonometric identity: sin²θ + cos²θ = 1. In this case, sinθ = √((v₁^2 + v₂^2) / v^2), and cosθ = √(1 - (v₁^2 + v₂^2) / v^2).
Therefore, we have:
sin²θ + cos²θ = 1
((v₁^2 + v₂^2) / v^2) + (1 - (v₁^2 + v₂^2) / v^2) = 1
(v₁^2 + v₂^2) / v^2 + 1 - (v₁^2 + v₂^2) / v^2 = 1
1 = 1
So, we see that sin²θ + cos²θ = 1, which means that the angle between the initial velocities of the objects is 90 degrees.