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.
30 minutes I am not sure about that
Answer:
A box sits stationary on a ramp
Explanation:
Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.
Static force of friction is calculated as follows:
F= μη
F is static force of friction.
μ is the coefficient of static friction.
η is the normal force.
Answer:
The statement is not correct.
Explanation:
To know if the statement is correct, we shall determine the velocity of the car after 3 s. This is illustrated below.
Data obtained from the question include:
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 3 s
Final velocity (v) =?
v = u + gt
v = 0 + (9.8 × 3)
v = 0 + 29.4
v = 29.4 m/s
Thus, the velocity of the car after 3 s is 29.4 m/s.
Hence, the statement made by the friend is not correct as the car has a falling velocity of 29.4 m/s after 3 s.
