Apparently, this website cannot display photos, so I'm gonna type my question:
A railway truck of mass 8340kg traveling at 14.3ms-1 collides with another truck of mass 6420kg traveling at 8.78ms-1 in the same direction.
1. If after collision the two trucks become joined together, what is their initial speed?
2.Calculate the percentage of kinetic energy retained in the collision.
please help I would be really grateful
To solve this problem, we'll use the principle of conservation of momentum and conservation of kinetic energy.
1. Let's start with the first question: "If after the collision, the two trucks become joined together, what is their initial speed?"
The principle of conservation of momentum tells us that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, we can express this as:
m1 * v1 + m2 * v2 = (m1 + m2) * v
Where m1 and m2 are the masses of the two trucks, v1 and v2 are their initial velocities, and v is their final velocity after collision.
Given:
m1 = 8340 kg (mass of truck 1)
v1 = 14.3 m/s (initial velocity of truck 1)
m2 = 6420 kg (mass of truck 2)
v2 = 8.78 m/s (initial velocity of truck 2)
Substituting these values into the equation, we have:
(8340 kg) * (14.3 m/s) + (6420 kg) * (8.78 m/s) = (8340 kg + 6420 kg) * v
Now we can solve for v:
v = [(8340 kg) * (14.3 m/s) + (6420 kg) * (8.78 m/s)] / (8340 kg + 6420 kg)
By evaluating this expression, we find that the final velocity (v) after the collision is approximately 12.22 m/s.
Therefore, the initial speed of the two trucks before the collision, when they become joined together, is 12.22 m/s.
2. Now let's move on to the second question: "Calculate the percentage of kinetic energy retained in the collision."
The principle of conservation of kinetic energy tells us that the total kinetic energy before the collision should be equal to the total kinetic energy after the collision. Mathematically, we can express this as:
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * (m1 + m2) * v^2
Using the same values for mass and velocity as given in the previous question, substitute them into the equation:
(1/2) * (8340 kg) * (14.3 m/s)^2 + (1/2) * (6420 kg) * (8.78 m/s)^2 = (1/2) * (8340 kg + 6420 kg) * (12.22 m/s)^2
By evaluating this expression, we find that the total kinetic energy after the collision is approximately 5.12 x 10^6 J.
To calculate the percentage of kinetic energy retained, we need to compare it with the initial kinetic energy. The initial kinetic energy can be calculated using the equation:
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2
Substituting the values, we get:
(1/2) * (8340 kg) * (14.3 m/s)^2 + (1/2) * (6420 kg) * (8.78 m/s)^2
By evaluating this expression, we find that the initial kinetic energy is approximately 6.77 x 10^6 J.
To find the percentage of kinetic energy retained, we divide the kinetic energy after the collision by the initial kinetic energy and multiply by 100:
(5.12 x 10^6 J / 6.77 x 10^6 J) * 100
By evaluating this expression, we find that approximately 75.6% of the initial kinetic energy is retained in the collision.
Therefore, the percentage of kinetic energy retained in the collision is approximately 75.6%.
8340(14.3)+6420(8.78) = (8340+6420)v
Ke initial = (1/2)(8340)(14.3^2) + (1/2)(8340)(8.78)^2
Ke final = (1/2)(8340+6420) v^2
it better be less.