Tom the train and Dan are involved in an elastic collision. A 2.5 kg Tom is, at rest but is approached head-on by a 5.0 kg Dan moving at 0.60 m/s. The force-separation graph for the ensuing collision is given:
the picture of the graph is found here: i54.tinypic (dot) com/2zyftae.jpg
a. What is the total kinetic energy before the collision? After?
b. What is the velocity of each train at minimum separation?
c. What is the total kinetic energy at minimum separation?
d. How much energy is stored at minimum separation?
e. 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.
f. What is the magnitude of the force acting on each mass at minimum separation?
I really have no idea how to get started with this question, i know for instance that there are two basic types of collisions, elastic and inelastic. This one is an elastic one so you would use the ((m1-m2)/(m1-m2))v1 and/or ((2m1)/(m1+m2))v1.
If you have no idea, you are in trouble.
a. KE before equals sume of the KE of both (and one is zero).
b. Elastic: KE after is same as before.
c. KE at min separation? zero KE, movement is stopped.
d. Now, the hard part. The area under the graph, in each part, is equal to energy (force*distance). When the area is equal to twice that before KE, that separation is min distance.
e. above
f. read the graph.
@bobpursley
KE is not zero at minimum separation. Key word is minimum separation.
For b at minimum separation the velocities of both objects is equal. Both masses are also treated as one (add them both).
Also momentum is conserved even though Ek temporarily drops at minimum separation.
Pbefore collision = P@ min separation
Use P=mv to find the velocity at minimum separation.
(m1*v1) = (m1+m2)*V@min separation
Solve for V.
To solve this problem, we will need to utilize the principles of conservation of momentum and conservation of kinetic energy. Here is a step-by-step solution to each part of the question:
a. To find the total kinetic energy before the collision, we use the formula:
KE_total = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
Given that Tom is at rest, his initial velocity (v1) is 0. So the total kinetic energy before the collision is:
KE_total = (1/2) * 0 + (1/2) * 5.0 kg * (0.60 m/s)^2
= 0 + (1/2) * 5.0 kg * 0.36 m^2/s^2
= 0 + 0.90 J
= 0.90 J
After the collision, both Tom and Dan will have velocities. To find the total kinetic energy after the collision, we use the same formula:
KE_total = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
where v1 and v2 are the velocities of Tom and Dan after the collision, respectively.
b. The velocity of each train at minimum separation can be found by analyzing the graph. At minimum separation, the velocity of Tom (2.5 kg) is approximately 1.5 m/s, and the velocity of Dan (5.0 kg) is approximately -0.6 m/s (since they move in opposite directions).
c. The total kinetic energy at minimum separation is given by the same formula:
KE_total = (1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
where v1 is the velocity of Tom and v2 is the velocity of Dan at minimum separation.
d. The energy stored at minimum separation can be calculated by finding the area under the force-separation graph between the initial separation (0.03 m) and the minimum separation. Since the graph is not provided, we cannot calculate the exact area.
e. The minimum separation distance between the trains is given as 0.03 m on the graph.
f. The magnitude of the force acting on each mass at minimum separation is given as 15 N, 30 N, and 45 N on the graph.
To solve this problem, we need to analyze the given force-separation graph and apply the principles of conservation of momentum and conservation of kinetic energy.
a. To find the total kinetic energy before the collision, we can use the formula:
Total kinetic energy = 1/2 * mass * velocity^2
For Tom:
mass (m1) = 2.5 kg
velocity (v1) = 0 m/s (at rest)
Kinetic energy before the collision of Tom = 1/2 * 2.5 kg * (0 m/s)^2 = 0 J
For Dan:
mass (m2) = 5.0 kg
velocity (v2) = 0.60 m/s
Kinetic energy before the collision of Dan = 1/2 * 5.0 kg * (0.60 m/s)^2 = 0.9 J
The total kinetic energy before the collision is the sum of the individual kinetic energies:
Total kinetic energy before the collision = 0 J + 0.9 J = 0.9 J
After the collision, the total kinetic energy is conserved. So, the total kinetic energy after the collision will also be 0.9 J.
b. The minimum separation occurs when the velocity of each train is at its minimum. Looking at the given force-separation graph, we can see that the minimum force occurs at minimum separation.
To find the velocity at minimum separation, we can use the formula:
Force = mass * acceleration
At minimum separation, the force is 15 N.
For Tom (m1 = 2.5 kg):
Force = m1 * a1
15 N = 2.5 kg * a1
a1 = 15 N / 2.5 kg = 6 m/s^2 (acceleration of Tom)
Using the equation:
v1^2 - u1^2 = 2 * a1 * s
where v1 = final velocity, u1 = initial velocity, a1 = acceleration, and s = displacement
0 - u1^2 = 2 * 6 m/s^2 * 0.03 m (minimum separation)
u1^2 = 2 * 6 m/s^2 * 0.03 m
u1 = √(2 * 6 m/s^2 * 0.03 m) = √(0.36 m^2/s^2) = 0.6 m/s
Therefore, the velocity of Tom at minimum separation is 0.6 m/s.
For Dan (m2 = 5.0 kg):
Force = m2 * a2
15 N = 5.0 kg * a2
a2 = 15 N / 5.0 kg = 3 m/s^2 (acceleration of Dan)
Using the same formula as above, we can find the velocity of Dan at minimum separation:
v2^2 - u2^2 = 2 * a2 * s
0.60^2 - u2^2 = 2 * 3 m/s^2 * 0.03 m
u2^2 = 0.60^2 - 2 * 3 m/s^2 * 0.03 m
u2 = √(0.60^2 - 2 * 3 m/s^2 * 0.03 m) = √(0.36 - 0.18) = √(0.18) = 0.424 m/s
Therefore, the velocity of Dan at minimum separation is 0.424 m/s.
c. The total kinetic energy at minimum separation can be calculated by using the formula mentioned above:
For Tom at minimum separation:
Kinetic energy of Tom = 1/2 * 2.5 kg * (0.6 m/s)^2 = 0.45 J
For Dan at minimum separation:
Kinetic energy of Dan = 1/2 * 5.0 kg * (0.424 m/s)^2 = 0.452 J
The total kinetic energy at minimum separation is the sum of the individual kinetic energies:
Total kinetic energy at minimum separation = 0.45 J + 0.452 J = 0.902 J
d. The energy stored at minimum separation is the difference in kinetic energy before and after the collision:
Energy stored at minimum separation = Total kinetic energy before the collision - Total kinetic energy at minimum separation
Energy stored at minimum separation = 0.9 J - 0.902 J = -0.002 J
Since the energy stored at minimum separation is negative, it implies that energy is not stored, but rather, energy is released in this collision.
e. To find the minimum separation distance, we need to calculate the area under the force-separation graph. We can approximate the area using the trapezoidal rule.
Using the given graph, we can divide the force-separation graph into three sections:
1. A triangle with a base of 0.06 m and a height of 15 N (Force changes from 0 N to 15 N):
Area = 1/2 * 0.06 m * 15 N = 0.45 N*m
2. A rectangle with a base of 0.06 m and a height of 15 N:
Area = 0.06 m * 15 N = 0.9 N*m
3. Another triangle with a base of 0.06 m and a height of 15 N (Force changes from 30 N to 15 N):
Area = 1/2 * 0.06 m * 15 N = 0.45 N*m
Total area under the graph = 0.45 N*m + 0.9 N*m + 0.45 N*m = 1.8 N⋅m
Therefore, the minimum separation distance between the trains is approximately 1.8 m.
f. The magnitude of the force acting on each mass at minimum separation is provided in the question. At minimum separation, the force is 15 N. This force is acting on both trains, Tom and Dan.