<u>Answer:</u>
The final velocity of the two railroad cars is 1.09 m/s
<u>Explanation:</u>
Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by
where
V= Final velocity
M1= mass of the first object in kgs = 12000
M2= mas of the second object in kgs = 10000
V1= initial velocity of the first object in m/s = 2m/s
V2= initial velocity of the second object in m/s = 0 (given at rest)
Substituting the given values in the formula we get
V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s
Which is the final velocity of the two railroad cars
Option(a) the mass of cart 2 is twice that of the mass of cart 1 is the right answer.
The mass of cart 2 is twice that of the mass of cart 1 is correct about the mass of cart 2.
Let's demonstrate the issue using variables:
Let,
m1=mass of cart 1
m2=mass of cart 2
v1 = velocity of cart 1 before collision
v2 = velocity of cart 2 before collision
v' = velocity of the carts after collision
Using the conservation of momentum for perfectly inelastic collisions:
m1v1 + m2v2 = (m1 + m2)v'
v2 = 0 because it is stationary
v' = 1/3*v1
m1v1 = (m1+m2)(1/3)(v1)
m1 = 1/3*m1 + 1/3*m2
1/3*m2 = m1 - 1/3*m1
1/3*m2 = 2/3*m1
m2 = 2m1
From this we can conclude that the mass of cart 2 is twice that of the mass of cart 1.
To learn more about inelastic collision visit:
brainly.com/question/14521843
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Answer:
Explanation:
If the work done on the cart is NET work
Then the work will result in an increase in kinetic energy
KE₀ + W = KE₁
½mv₀² + W = ½mv₁²
½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²
v₁ = 1.626991...
v₁ = 1.6 m/s
A)Ep'=mgh=mgl(1-cosa).At the bottom of the swing Ep=0(reference level),so the potential energy as the child is just released is bigger than the potential energy at the bottom of the swing.;B)The speed of the child at the bottom of the swing-->v=√(2gh)=√[2gl(1-cosa)];C)I don't think that the tension does any work.
Answer:
it started to move a 1 second
