Maybe you can split up the questions. I will try to answer your first question.
1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)
2. Momentum: p = m₁v₁ + m₂v₂
m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B
Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0
Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0
3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.
4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7
5.You figure out.
Answer:
17.71N/m
Explanation:
The period of the spring is expressed according to the expression;
m is the mass of the object
k is the force constant
Given
m = 5.50kg
T = 3.50s
Substitute into the formula;
Hence the force constant of the spring is 17.71N/m
Its part of the electromagnetic spectrum. it works by emitting from the sun.
if thats not it let me know more specificly what you are doing please
Answer:
A. The momentum of car A(5kg) is EQUAL to that of car B(0.5)
Explanation:
The moment, or impulse formula of the same forces acting on both car within 1 second is
In our case the forces are the same, the time duration of force acting on the cars are the same. Therefore, their momentum right after the force must also be the same.
