The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the t
1 answer:
Answer:
The speed of the boxes are 1 m/s.
Explanation:
Given that,
Mass of box = 1 kg
Mass of another box = 2 kg
Suppose 1 kg box moves with 3 m/s speed.
We need to calculate the speed of the boxes
Using formula of conservation of momentum
Where, u = initial velocity
v = final velocity
Put the value into the formula
Hence, The speed of the boxes are 1 m/s.
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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))
