Thomas the train and Diesel are involved in an elastic collision. A 2.5 kg Thomas is, at rest but is approached head-on by a 5.0 kg Diesel moving at 0.60 m/s. The force-separation graph for the ensuing collision is given. (12 marks)
What is the total kinetic energy before the collision? After?
What is the velocity of each train at minimum separation?
What is the total kinetic energy at minimum separation?
How much energy is stored at minimum separation?
What is the minimum separation distance between the trains? Hint: The energy temporarily stored at minimum separation equals a portion of the area under the above graph. The collision starts when the centers of the trains are separated by 0.03 m as shown on the above graph at which time the collision force is 15 N. But this force increases to 30 N and then eventually 45 N.
What is the magnitude of the force acting on each mass at minimum separation?
To answer these questions, we need to analyze the given force-separation graph and apply the principles of conservation of energy and momentum.
1. Total kinetic energy before the collision: Since Thomas is initially at rest, its kinetic energy is zero. Diesel has a mass of 5.0 kg and a velocity of 0.60 m/s. The formula for kinetic energy is KE = 0.5 * mass * velocity^2. Plugging in the values, we have KE = 0.5 * 5.0 kg * (0.60 m/s)^2 = 0.9 J. So, the total kinetic energy before the collision is 0.9 J.
2. Total kinetic energy after the collision: In an elastic collision, kinetic energy is conserved. Since there is no loss of kinetic energy, the total kinetic energy after the collision will also be 0.9 J.
3. Velocity of each train at minimum separation: At minimum separation, the forces applied to both trains are equal. According to Newton's third law of motion, if the force on Diesel is pushing it back, the same force will be acting on Thomas in the opposite direction. Therefore, the velocities of Thomas and Diesel at minimum separation will be equal in magnitude but opposite in direction. Let's assume the velocity is v. Using the principle of conservation of momentum, we have:
(2.5 kg * 0 m/s) + (5.0 kg * 0.60 m/s) = (2.5 kg * (-v)) + (5.0 kg * v)
Solving this equation, we find v = -0.20 m/s. Therefore, Thomas and Diesel have velocities of -0.20 m/s and 0.20 m/s, respectively, at minimum separation.
4. Total kinetic energy at minimum separation: Using the same formula as before, the kinetic energy of Thomas at minimum separation is KE = 0.5 * 2.5 kg * (-0.20 m/s)^2 = 0.02 J. The kinetic energy of Diesel at minimum separation is KE = 0.5 * 5.0 kg * (0.20 m/s)^2 = 0.10 J. So, the total kinetic energy at minimum separation is 0.02 J + 0.10 J = 0.12 J.
5. Energy stored at minimum separation: The energy temporarily stored at minimum separation is equal to the area under the force-separation graph from the start of the collision (0.03 m) up to the minimum separation. This can be calculated by finding the area of the triangle formed by the base and height. The base is the difference in separation distances (0.03 m - 0 m = 0.03 m) and the height is the average of the forces (15 N + 30 N) / 2 = 22.5 N.
Energy stored = 0.5 * base * height
= 0.5 * (0.03 m) * (22.5 N)
= 0.3375 J
Therefore, the energy stored at minimum separation is 0.3375 J.
6. Minimum separation distance between the trains: The minimum separation distance between the trains is given as 0.03 m.
7. Magnitude of the force acting on each mass at minimum separation: According to the graph, the force at minimum separation is given as 15 N. Since the forces on both trains are equal, the magnitude of the force acting on each mass at minimum separation is 15 N.