Is it possible for the momentum of a system consisting of two carts on a low-friction track to be zero even if both carts are mo
1 answer:
Answer:
Yes, if the two carts are moving into opposite directions
Explanation:
The total momentum of the system of two carts is given by:
where
m1, m2 are the masses of the two carts
v1, v2 are the velocities of the two carts
Let's remind that v (the velocity) is a vector, so its sign depends on the direction in which the cart is moving.
We want to know if it is possible that the total momentum of the system can be zero, so it must be:
From this equation, we see that this condition can only occur if v1 and v2 have opposite signs. Opposite signs mean opposite directions: therefore, the total momentum can be zero if the two carts are moving into opposite directions.
You might be interested in
Answer:
C = 4,174 10³ V / m^{3/4} , E = 7.19 10² / ∛x, E = 1.5 10³ N/C
Explanation:
For this exercise we can calculate the value of the constant and the electric field produced,
Let's start by calculating the value of the constant C
V = C
C = V / x^{4/3}
C = 220 / (11 10⁻²)^{4/3}
C = 4,174 10³ V / m^{3/4}
To calculate the electric field we use the expression
V = E dx
E = dx / V
E = ∫ dx / C x^{4/3}
E = 1 / C x^{-1/3} / (- 1/3)
E = 1 / C (-3 / x^{1/3})
We evaluate from the lower limit x = 0 E = E₀ = 0 to the upper limit x = x, E = E
E = 3 / C (0- (-1 / x^{1/3}))
E = 3 / 4,174 10³ (1 / x^{1/3})
E = 7.19 10² / ∛x
for x = 0.110 cm
E = 7.19 10² /∛0.11
E = 1.5 10³ N/C
Rearranging formulas is all about simple algebra rules. Just like when solving for x in an equation, you're just isolating whichever variable you want. I'll work this one out for you and hopefully it'll help, but if you need more explanation, then feel free to comment!
D = ViT + 0.5at² Subtract ViT from both sides
D - ViT = 0.5at² Divide both sides by 0.5t²
I wrote this step out a little more to show how your fraction will cancel
= a I like to flip these around so the single variable is on the right
a =