The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:
m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,
m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂' --> equation 1
The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is
(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2
Solving equations 1 and 2 simultaneously, v₁' = -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
Answer:
W = - 5.01 10¹⁰ J
Explanation:
Work is defined by the expression
W = ∫ F.dr
Where the blacks indicate vectors, in the case the force is radial and the distance is also radial, whereby the scalar producer is reduced to an ordinary product
W = ∫ F dr
W = G m₁m₂ ∫ 1 /r² dr
W = G m₁ m₂2(-1 / r)
We evaluate between the lower limits r = Re and upper r = ∞
W = G m₁m₂ (-1 / Re + 1 / ∞)
W = - G m₁ m₂ / Re
Let's calculate
W = - 6.67 10⁻¹¹ 800 5.98 10²⁴ / 6.37 10⁶
W = - 5.01 10¹⁰ J
Answer:
D) 15s
Explanation:
let Te be the period of the block-spring system on earth and Tm be the period of the same system on the moon.let g1 be the gravitational acceleration on earth and g2 be the gravitational acceleration on the moon.
the period of a pendulum is given by:
T = 2π√(L/g)
so on earth:
Te = 2π√(L/g1)
= 6s
on the moon;
Tm = 2π√(L/g2)
since g2 = 1/6 g1 then:
Tm = 2π√(L/(1/6×g1))
= √(6)×2π√(L/(g1))
and 2π√(L/(g1)) = Te = 6s
Tm = (√(6))×6 = 14.7s ≈ 15s
Therefore, the period of the block-spring system on the moon is 15s.