A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.80
1 answer:
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Answer:
(a) The speed of the lighter block is .
The speed of the heavier block is .
(b) The smaller block goes up to 0.69 m.
Explanation:
We will divide this question into three parts: Part A is for the smaller mass from the top of the incline to the collision. Part B is the collision. And Part C is for the smaller mass from the bottom to the highest point it can achieve.
In order to solve this question, I will assume that the smaller mass is initially at rest.
Part A:
We will use conservation of energy.
This is the speed of the smaller mass just before the collision. The velocity of the mass is directed 30° above horizontal, since the mass is sliding down the incline.
Part B:
Momentum is a vector identity, so the x- and y-components of momentum are to be investigated separately. Since the collision occurs at the horizontal surface, only the x-component of momentum is conserved.
During the collision kinetic energy is also conserved. Since kinetic energy is a scalar quantity, we don't have to separate its components.
The following relation will be used when combining the two equations:
The following equation is useful for combining the two equations:
Therefore from the first equation,
Part C:
We will again use the conservation of energy to find the highest point that the mass can go:
Answer:
k =
b =
t =
Solution:
As per the question:
Mass of the block, m = 1000 kg
Height, h = 10 m
Equilibrium position, x = 0.2 m
Now,
The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:
where
u = initial velocity = 0
g = 10
Thus
Force on the mass is given by:
F = mg =
Also, we know that the spring force is given by:
F = - kx
Thus
Now, to find the damping constant b, we know that:
F = - bv
Now,
Time required for the platform to get settled to 1 mm or 0.001 m is given by: