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:
Explanation:
We have to find electric potential V at a distance r.
a) For r>R,
The electric field in the cylinder is given by
E.A equating it to the other electric field given by
б.A/ε₀
Here the area of cylinder is given by= 2*3.14*r*L
While for the outside, the area= 2*3.14*R*L
Equating both, we get
E= бR/rε₀
Now,
The potential difference is given as:
ΔV= -бR/rε₀ and integrating right side with respect to dr under limits r and R.
Where ΔV= V₀-V
So solving we get
V₀=V-бR/ε₀ln (r/R)
b) For r<R i.e. inside the cylinder
There will be no electric field produced as E=0
So ultimately Vin= V
c) V=0 at r= infinity.
Assuming that reaching a height 0 doesn’t stop the ball, and that it accelerates at 9.8 m/s^2, the ball would be traveling at 0.5 + 0.7*9.8 = 7.36 m/s downwards.
