Momentum would be the same before and after the collision
Before the collision:
Momentum of the single cart: 1 * 0.50 = 0.50
After the collision
velocity = 0.25m / s
1 * 0.25 + 1 * 0.25 =
0.25 * (1 + 1) =
0.25 * 2 =
0.50
Now new momentum will be 0.5
answer
the same before and after the collision
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
Answer:
0.010 m
Explanation:
So the equation for a pendulum period is:
where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:

Evaluate the multiplication in front

Divide both sides by 6.28

Square both sides

Multiply both sides by m/s^2 (the s^2 will cancel out)
Now now let's find the length when it's two seconds

Divide both sides by 6.28

Square both sides

Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)

So to find the difference you simply subtract
0.984 - 0.994 = 0.010 m
Answer:
Explanation:I don't say you have to mark my ans as brainliest but if you think it has really helped you plz don't forget to thank me ...