At which angles relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344 m/s
1 answer:
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a) 6.4 m/s
b) 2.1 m
c)
d) 14.0 N
e) 4.6 m/s
f) 37.9 N
Explanation:
a)
Since the system is isolated (no external forces on it), the total momentum of the system is conserved, so we can write:
where:
is the mass of the 1st sphere
is the mass of the 2nd sphere
is the initial velocity of the 1st sphere
is the final velocity of the 1st sphere
is the final velocity of the 2nd sphere
Since the collision is elastic, the total kinetic energy is also conserved:
Combining the two equations together, we can find the final velocity of the 2nd sphere:
b)
Now we analyze the 2nd sphere from the moment it starts its motion till the moment it reaches the maximum height.
Since its total mechanical energy is conserved, its initial kinetic energy is entirely converted into gravitational potential energy at the highest point.
So we can write:
where
m = 3 kg is the mass of the sphere
v = 6.4 m/s is the initial speed of the sphere
is the acceleration due to gravity
h is the maximum height reached
Solving for h, we find
c)
Here the 2nd sphere is tied to a rope of length
L = 4 m
We know that the maximum height reached by the sphere in its motion is
h = 2.1 m
Calling the angle that the rope makes with the vertical, we can write
Which can be rewritten as
Solving for , we can find the angle between the rope and the vertical:
d)
The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:
- The tension in the rope, T, inward
- The component of the weight along the radial direction, , outward
Their resultant must be equal to the centripetal force, so we can write:
where r = L (the radius of the circle is the length of the rope).
However, when the sphere is at the highest point, it is at rest, so
v = 0
Therefore we have
So we can find the tension:
e)
We can solve this part by applying again the law of conservation of energy.
In fact, when the sphere is at a height of h = 1 m, it has both kinetic and potential energy. So we can write:
where:
is the initial kinetic energy
is the kinetic energy at 1 m
is the final potential energy
v = 6.4 m/s is the speed at the bottom
v' is the speed at a height of 1 m
h' = 1 m is the height
m = 3 kg is the mass of the sphere
And solving for v', we find:
f)
Again, since the sphere is in circular motion, the equation of the forces along the radial direction is
where
T is the tension in the string
is the component of the weight in the radial direction
is the centripetal force
In this situation we have
v = 4.6 m/s is the speed of the sphere
can be rewritten as (see part c)
where in this case,
h' = 1 m
L = 4 m
And is the radius of the circle
Substituting and solving for T, we find: