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Featured 2 months ago

Velocity of 3 kg ball: v₁ = 2.8 m s⁻¹ to the left

Velocity of 7 kg ball: v₂ = 2.3 m s⁻¹ to the right

m₁ = 3 kg

u₁ = 5 m s⁻¹

m₂ = 7 kg

u₂ = -1 m s⁻¹

Conservation of momentum:

Substitute values into equation:

Conservation of energy:

If 25% of the kinetic energy is lost then the final kinetic energy is equal to 75% of the initial kinetic energy.

⇒

Substitute values into equation:

**Solve the simultaneous equations ① and ②:**

From ①:

Substitute for v₁ in equation ②:

⇒

⇒

Use an equation to solve for the factors of the quadratic, i.e. this will give possible solutions for v₂.

Solution 1:

Solution 2:

Substitute each of those solutions into equation 1 to find possible solutions for v₁:

E.g.

**Now interpret the results qualitatively (in words):**

Solution set 1 (v₁ = -2.786 m s⁻¹ and v₂ = 2.337 m s⁻¹) : ball 1 reverses direction at a lower velocity and ball 2 reverses direction at a faster velocity.

Solution set 2 (v₁ = 4.386 m s⁻¹ and v₂ = -0.7367 m s⁻¹) : ball 1 slows down slightly and ball 2 also slows down slightly, they both continue in the same direction.

Whilst the two solutions are mathematically satisfactory we can see from a qualitative analysis that solution two is physically impossible. The two balls were initially moving toward each other, they cannot both just slow down slightly after colliding and continue in the same direction *unless they magically pass through each other*.

**For that reason the solution to the problem is that the final velocities are:**

v₁ = -2.8 m s⁻¹ (i.e. 2.8 m s⁻¹ to the left)

v₂ = 2.3 m s⁻¹ (2.3 m s⁻¹ to the right)

Featured 2 months ago

Considering

For

Horizontally

Vertically

here

For

Horizontally

Vertically

here

Now the global equilibrium states that the momentum of the forces acting over the tube must sum zero or

Here we suppose that the tube has the right bottom extremum as possible rotational pivot, at the knock down limit, so the support normal forces are concentrated at this point.

Here

Solving the system of equations

for

so

d'Alembert's principle analysis will come later.

Analysing the virtual work of each body

Now calling

Finally

Note:

The detailed calculation of

After a rigid rotation of

then

but

and also

Finally

Featured 2 months ago

The Law of Conservation of Energy is one of the most important Physic's Laws. This law says that energy never is destroyed or created, it just changes from one form of energy to another.

Without going into details of the latest theories on the structure of the Universe and in particular theories about space, time and origin of matter, the principle of conservation of energy is a universal law.

This law says that the energy of the universe remains constant. This means that in any physical or chemical process, the total energy of universe remains constant. If a form of energy increases or decreases is always at the expense of other forms of energy decreases or increases, respectively.

A practical example of this would be the case of a body falling from a height. Suppose we are at a height

What has happened from the point of view of energy?

When the object was at a height

We release the object and this is accelerating as it falls. On reaching the floor height is zero, so that the potential energy is now

If

Featured 1 month ago

Sounds like efficiency to me. In general, efficiency is defined as the work output acquired from some work input.

where:

#w# is work.#q_H# is the heat flow into the engine from the hot reservoir.#q_C# is the heat flow from the engine into the cold reservoir.#T_H# is the temperature of the hot reservoir.#T_C# is the temperature of the cold reservoir.

The **ideal** machine **can** achieve **Real** machines, however, can **never quite achieve** *nearly*

As I derived above, a cyclic process has a change in entropy of

If we separate the cyclic process into two processes, then

#q_H/T_H + q_C/T_C = 0#

Also from the above, I calculated that

#color(blue)(e) = |w|/q_H = (q_H + q_C)/q_H = color(blue)(1 + q_C/q_H)# .

Or, using the first relation:

#color(blue)(e) = |w|/q_H = (q_H + q_C)/q_H = color(blue)(1 - T_C/T_H)# .

So, we could even say that since all absolute temperatures are positive (i.e. when in

Therefore, the efficiency can never be more than

Featured 1 week ago

The speed of the astronaut changes by

This is a problem of momentum conservation, specifically an explosion. In an explosion, an internal impulse acts in order to propel the parts of a system into a variety of directions. This is often a single object, but can be, such as in this case, multiple objects which were initially at rest together. For collisions/explosions occurring in an isolated system, momentum is always conserved-no exceptions.

The Law of Conservation of Momentum:

#vecP_f=vecP_i#

For multiple objects,

#vecP=vecp_(t ot)=sumvecp=vecp_1+vecp_2+...vecp_n#

In our case, we have the momentum of the thrown object and the astronaut. We'll call the mass of astronaut

#m_1v_(1i)+m_2v_(2i)=m_1v_(1f)+m_2v_(2f)#

However, both the thrown object and the astronaut are initially at rest, so the total momentum before the "explosion" is

#0=m_1v_(1f)+m_2v_(2f)#

We can manipulate the equation to solve for

#v_(f1)=-(m_2v_(f2))/m_1#

Using our known values:

#v_(f1)=-((16kg)(3/8m/s))/(80kg)#

#=>v_(f1)=-3/40m/s=-0.075m/s#

In the above equations we've defined the direction that the object is thrown in as positive, so a negative answer tells us that astronaut moves in the opposite direction of the object. Because his initial velocity was

Hope this helps!

Featured 1 week ago

The ball will reach a maximum altitude of

Ignoring air resistance, friction, etc., this problem can be solved with energy conservation. I will use the explanation I gave for a similar question in the past. Find **tl;dr** to skip mini lesson in energy conservation.

#E_(mech)=K+U#

Where

#E_i=E_f#

#K_i+U_i=K_f+U_f#

From this relationship we can see that if potential energy, for example, were to decrease, kinetic energy would have to increase in order for energy to be conserved. The inverse is also true. The idea is that energy is being transformed between two forms (potential and kinetic) rather than any of it being lost of gained. When we have to include friction in problems, for example, energy is not conserved, because some of it is lost to heat (

Initially, assuming the spring is compressed to point where it's height is insignificant (so that we need not worry about gravitational potential energy of the ball beforehand), all energy is stored as spring potential energy in the spring. Because nothing is moving prior to the launch, there is no kinetic energy. Therefore, we have only potential energy to begin with.

#U_(sp)=U_f+K_f#

After the launch, as a the spring decompresses, the energy stored within it as spring potential energy is transferred to the ball, which is observed as the ball shoots upward. The spring potential energy has been transformed into kinetic energy. However, as the ball rises, it is gaining altitude, and therefore gaining gravitational potential energy. Because the potential energy of the ball increases, its kinetic energy must decrease, and we observe this in the form of the ball stopping at some point in the air before falling back to the earth. This is the point at which all of the kinetic energy has been transformed into gravitational potential energy.

Because we are concerned with the maximum altitude of the ball, what we're really asking is at what point all of the potential energy from the spring has been transferred into gravitational potential energy, which is given by

#U_(sp)=U_g#

Spring potential energy,

#1/2k(Δs)^2=mgh#

( **tl;dr** ) Rearranging to solve for

#h=(1/2k(Δs)^2)/(mg)#

Using our known values for mass, the spring constant, and the compression of the spring, we can now calculate a value for

#h=(1/2(28N/m)(9/6m)^2)/(0.420kg*9.8m/s^2)#

#h~~5.1m#

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