The chemical equation needs to be balanced so that it follows the law of conservation of mass. A balanced chemical equation occurs when the number of the different atoms of elements in the reactants side is equal to that of the products side. Balancing chemical equations is a process of trial and error.
1) In any collision the momentum is conserved
(2*m)*(vo) + (m)*(-2*vo) = (2*m)(v1') + (m)(v2')
candel all the m factors (because they appear in all the terms on both sides of the equation)
2(vo) - 2(vo) = 2(v1') + (v2') => 2(v1') + v(2') = 0 => (v2') = - 2(v1')
2) Elastic collision => conservation of energy
=> [1/2] (2*m) (vo)^2 + [1/2](m)*(2*vo)^2 = [1/2](2*m)(v1')^2 + [1/2](m)(v2')^2
cancel all the 1/2 and m factors =>
2(vo)^2 + 4(vo)^2 = 2(v1')^2 + (v2')^2 =>
4(vo)^2 = 2(v1')^2 + (v2')^2
now replace (v2') = -2(v1')
=> 4(vo)^2 = 2(v1')^2 + [-2(v1')]^2 = 2(v1')^2 + 4(v1')^2 = 6(v1')^2 =>
(v1')^2 = [4/6] (vo)^2 =>
(v1')^2 = [2/3] (vo)^2 =>
(v1') = [√(2/3)]*(vo)
Answer: (v1') = [√(2/3)]*(vo)
Answer:
(a) Position Vectors V₁= -2î km, V₂=5î km
(b) Displacement Δx=7 km
Explanation:
Given data
Distance=2 km west at t=0
Distance=5 km east at t=6 min
Positive x is the east direction
To find
(a)Car position vector at given times
(b)Displacement between 0 to 6.0 min
Solution
For Part (a) car position vector at given times
At t=0 the distance=2 km west so conclude that x₁=-2 because it is in negative side So vector V₁
V₁= -2î km
At t=6.0 the distance=5 km east so conclude that x₂=5 because it is in positive side So vector V₂
V₂=5î km
For (b) displacement between 0 to 6.0 min
According to following mathematical law we can conclude that
Δx=x₂-x₁
Δx=5-(-2)km
Δx=7 km
Even though the mass of the earth stays the same, the force gravity has on us will double because there is less space in the atmosphere for the gravity to pull anything in towards its center. Let me know if I'm correct!