3. A train loaded with steel, moving at 90km/h, collides head on with a stationary train of mass 6000kg. if the trains couple after collision and move forward with a velocity of 80km/h, find the mass of the train loaded with steel.
My work so far:
PTO=PTF
mivi+m2v2=mvf
mi (25m/s)=(mi + 6000kg)(22.2m/s)
and then I get stuck here because I find it virtually impossible to solve for mi. I know the answer is suppose to be 4.8x10^4 kg though.
V1 = 90km/h = 90000m /3600s = 25 m/s.
V = 80km/h = 80000m/3600s = 22.2 m/s.
M1*V1 + M2*V2 = M1*V + M2*V.
M1*25+6000*0=M1*22.2 + 6000*22.2
25M1-22.2M1 = 6000*22.2.
2.78M1 = 133,200.
M1 = 47,952 kg. = 4.8*10^4 kg.
Well, solving this problem may feel as tricky as trying to untangle a bunch of clowns in a tiny car, but fear not! I'll try to help you through it with a touch of humor.
Let's break it down step by step, or should I say, car by car:
1. Convert the given velocities into meters per second:
The initial velocity of the loaded train is 90 km/h, which is approximately 25 m/s.
The final velocity of the coupled trains is 80 km/h, which is approximately 22.2 m/s.
2. Apply the conservation of momentum:
We have mi * 25 = (mi + 6000) * 22.2.
3. Expand and simplify:
25mi = 22.2mi + 132,000.
Subtract 22.2mi from both sides:
2.8mi = 132,000.
Divide both sides by 2.8:
mi = 47,142.857142857...
4. Well, that's quite a specific mass! But since we're dealing with trains and precision, let's round it to the nearest whole number:
The mass of the train loaded with steel is approximately 47,143 kg.
However, the answer you mentioned, 4.8x10^4 kg, is equivalent to 48,000 kg. So, there appears to be a discrepancy. Please double-check your calculations and let me know if you have any other questions. Remember, math can be as unpredictable as a circus, but we'll get through it together!
To find the mass of the train loaded with steel, let's break down the calculation step-by-step:
1. Given information:
- Mass of the stationary train (m2) = 6000 kg
- Initial velocity of the loaded train (vi) = 90 km/h = 25 m/s
- Final velocity of the coupled trains (vf) = 80 km/h = 22.2 m/s
2. Apply the principle of conservation of momentum:
According to the principle of conservation of momentum, the total momentum before the collision (PTO) is equal to the total momentum after the collision (PTF).
PTO = PTF
(m1 * vi) + (m2 * 0) = (m1 + m2) * vf
Since the second train is stationary, its initial velocity is 0.
3. Substitute the given values into the equation:
(m1 * 25) + (6000 * 0) = (m1 + 6000) * 22.2
This simplifies to:
25m1 = 22.2m1 + 133,200
2.8m1 = 133,200
m1 = 133,200 / 2.8
m1 = 47,571.43 kg
Hence, the mass of the train loaded with steel is approximately 47,571 kg or 4.8 × 10^4 kg.
To solve this problem, you can apply 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. The momentum of an object is given by the product of its mass and velocity.
Let's assign variables to the given quantities:
Mass of the stationary train = m1 = 6000 kg
Velocity of the loaded train before the collision = v1 = 90 km/h = 25 m/s
Velocity of the loaded train after the collision = vf = 80 km/h = 22.2 m/s
Using the conservation of momentum equation, we have:
m1v1 + m2v2 = (m1 + m2)vf
Substituting the given values:
6000 kg * 25 m/s + m2 * 0 m/s = (6000 kg + m2) * 22.2 m/s
Now, solve this equation for m2:
150000 kg⋅m/s + 0 m2 = 133200 kg⋅m/s + 22.2m2
Rearrange the terms:
-22.2 m2 + 22.2 m2 = 133200 kg⋅m/s - 150000 kg⋅m/s
-22.2 m2 = -16800 kg⋅m/s
m2 = (-16800 kg⋅m/s) / (-22.2 m/s)
m2 ≈ 757.58 kg
Therefore, the mass of the train loaded with steel is approximately 757.58 kg.
It appears there might be an error in the given answer of 4.8x10^4 kg, as the calculated value is significantly lower. Please double-check the question or consider recalculating the answer.