A low friction cart of mass m, traveling at speed v, collides (perfectly) elastically with a cart of triple the
mass, initially at rest. What is the result of this collision? (What are the final, after-collision velocities?) Do
NOT use the derived equations from the textbook. Work this out and show the work.
I will be happy to try to check your work.
You have two equations: conservation of momentum, and energy.
Bob, AKA smart guy,
I already figured it out thank you very much. You obviously didn't read the question as it says don't use derived equations from the textbook. Wouldn't you assume conservation of energy and momentum equations are in the book?
To find the final velocities of the carts after the elastic collision, we can use the principle of conservation of momentum.
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Let's denote the mass of the low friction cart as m₁ and the mass of the other cart as m₂. The initial velocity of the low friction cart is v₁ and the initial velocity of the other cart is 0 (since it is initially at rest). Let's denote the final velocities of the two carts as v₁' and v₂'.
The momentum of an object is given by the product of its mass and velocity:
Momentum before collision = m₁ * v₁ + m₂ * 0 (since the other cart is initially at rest)
Momentum after collision = m₁ * v₁' + m₂ * v₂'
Now, since the collision is perfectly elastic, 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 the equation KE = (1/2) * m * v².
Therefore, we can write:
(1/2) * m₁ * v₁² = (1/2) * m₁ * v₁'² + (1/2) * m₂ * v₂'²
Now, to solve for the unknown velocities v₁' and v₂', we need another equation. This second equation comes from the conservation of momentum:
m₁ * v₁ = m₁ * v₁' + m₂ * v₂'
Now, we have a system of two simultaneous equations:
(1) (1/2) * m₁ * v₁² = (1/2) * m₁ * v₁'² + (1/2) * m₂ * v₂'²
(2) m₁ * v₁ = m₁ * v₁' + m₂ * v₂'
We can solve these two equations simultaneously to find the values of v₁' and v₂'. Let's rearrange equation (2) to solve for v₁':
v₁ = v₁' + (m₂ / m₁) * v₂'
Now, substitute this expression for v₁' in equation (1):
(1/2) * m₁ * v₁² = (1/2) * m₁ * [v₁ - (m₂ / m₁) * v₂']² + (1/2) * m₂ * v₂'²
Simplify and solve this equation algebraically for v₂':
(1/2) * m₁ * v₁² = (1/2) * m₁ * (v₁² - 2 * (m₂ / m₁) * v₁ * v₂' + (m₂ / m₁)² * v₂'²) + (1/2) * m₂ * v₂'²
Expand the equation further:
1/2 * m₁ * v₁² = 1/2 * m₁ * v₁² - m₂ * v₁ * v₂' + 1/2 * m₂ * (m₂ / m₁)² * v₂'² + 1/2 * m₂ * v₂'²
Cancel out the common terms:
0 = - m₂ * v₁ * v₂' + 1/2 * m₂ * (m₂ / m₁)² * v₂'² + 1/2 * m₂ * v₂'²
Combine like terms:
m₂ * v₁ * v₂' = 1/2 * m₂ * (m₂ / m₁)² * v₂'² + 1/2 * m₂ * v₂'²
As can be seen, this equation includes a quadratic term (v₂'²), so we can solve it using appropriate algebraic techniques such as factorization or the quadratic formula.
Upon solving the quadratic equation, we can find the value(s) of v₂' (the final velocity of the other cart) in terms of the given variables (m₁, m₂, v₁).
Once we have obtained the value of v₂', we can now substitute it back into equation (2) to find v₁'.
Finally, we will have the final velocities of both carts after the elastic collision.