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A ball with a mass of #3 kg # and velocity of #5 m/s# collides with a second ball with a mass of #7 kg# and velocity of #- 1 m/s#. If #25%# of the kinetic energy is lost, what are the final velocities of the balls?

Daniel W.
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

Explanation:

m₁ = 3 kg
u₁ = 5 m s⁻¹
m₂ = 7 kg
u₂ = -1 m s⁻¹

Conservation of momentum:
#m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂#
Substitute values into equation:
#3 × 5 + 7 × (-1) = 3 v₁ + 7 v₂ #
#⇒ 8 = 3 v₁ + 7 v₂ # [equation ① ]

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.
#0.75*(½m_1 u_1^2 + ½m_2 u_2^2) = ½m_1 v_1^2 + ½m_2 v_2^2#
$0.75 \cdot \left({m}_{1} {u}_{1}^{2} + {m}_{2} {u}_{2}^{2}\right) = {m}_{1} {v}_{1}^{2} + {m}_{2} {v}_{2}^{2}$
Substitute values into equation:
#0.75(3 × 25 + 7 × 1) = 3 v_1^2 7 v_2^2#
#⇒ 3 v_1^2 7 v_2^2 = 61.5# [equation ② ]

Solve the simultaneous equations ① and ②:
From ①: # v₁ = (8 - 7 v₂)/3#
Substitute for v₁ in equation ②:
$3 {\left(\frac{8 - 7 {v}_{2}}{3}\right)}^{2} 7 {v}_{2}^{2} = 61.5$
$\frac{1}{3} \left(8 - 7 {v}_{2}\right) \left(8 - 7 {v}_{2}\right) + 7 {v}_{2}^{2}$
$23.33 {v}_{2}^{2} - 37.33 {v}_{2} - 40.17 = 0$

Use an equation to solve for the factors of the quadratic, i.e. this will give possible solutions for v₂.
#v_2 = (-B ± sqrt(B^2- 4AC))/2A#

Solution 1: #v_(2₁) = 2.337 m s⁻¹#
Solution 2: #v_(2₂) = -0.7368 m s⁻¹#

Substitute each of those solutions into equation 1 to find possible solutions for v₁:
E.g. #v_(1₁) = (8 - 7 × 2.337)/3 = -2.786 m s⁻¹#
#v_(1₂) = 4.386 m s⁻¹#

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)

Can you solve this problem using Newton mechanics and proving the result with D'Alembert's principle? I have no idea

Cesareo R.
Featured 2 months ago

$M = 2 m \left(1 - \frac{r}{R}\right)$

Explanation:

Considering ${B}_{1}$ and ${B}_{2}$ with ${B}_{1}$ at the botton, its static equilibrium is attained with the forces

For ${B}_{1}$

Horizontally

${f}_{1} - {f}_{c} \cos \theta = 0$

Vertically

$- m g + {n}_{1} - {f}_{c} \sin \theta = 0$

here ${f}_{1}$ is the horizontal contact force with the tube wall and ${f}_{c}$ is the contact force between ${B}_{1}$ and ${B}_{2}$ and ${n}_{1}$ the floor contact force.

For ${B}_{2}$

Horizontally

${f}_{c} \cos \theta - {f}_{2} = 0$

Vertically

$- m g + {f}_{c} \sin \theta = 0$

here ${f}_{2}$ is the horizontal contact force with the tube ball.

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

${f}_{1} r - {f}_{2} \left(r + 2 r \sin \theta\right) + M g R = 0$

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 $\cos \theta = \frac{R - r}{r}$ and $\sin \theta = \frac{\sqrt{2 r R - {R}^{2}}}{r}$

Solving the system of equations

#{ (f_1-f_c cos theta=0), (-m g + n_1 -f_c sintheta=0), (f_c costheta-f_2=0), (-m g +f_c sintheta = 0), (f_1 r-f_2(r+2r sintheta)+M g R =0) :}#

for ${f}_{1} , {f}_{2} , {f}_{c} , {n}_{1} , M$ we obtain

#f_1 = (costheta g m)/sintheta, f_2 = (costheta g m)/sintheta, f_c = (g m)/sintheta, n_1 = 2 g m, M = (2 costheta m r)/R#

so

$M = 2 m \left(1 - \frac{r}{R}\right)$

d'Alembert's principle analysis will come later.

Analysing the virtual work of each body ${B}_{1} , {B}_{2}$ and $T$ (for tube) we have $\delta w = \left\langle\delta p , f\right\rangle$

$\delta {p}_{1} = r \delta \alpha \hat{i}$

$\delta {p}_{2} = \delta {p}_{1} + 2 r \left(\cos \left(\theta - \delta \alpha\right) - \cos \left(\theta\right)\right) \hat{i} + 2 r \left(\sin \left(\theta - \delta \alpha\right) - \sin \left(\theta\right)\right) \hat{j}$ so

$\delta {p}_{2} \approx r \delta \alpha \hat{i} + 2 r \sin \theta \delta \alpha \hat{i} - 2 r \cos \theta \delta \alpha \hat{j}$

Now calling ${p}_{T} = l \cos \alpha \hat{i} + l \sin \alpha \hat{j}$ where $l$ is the distance between the pivot point and the tube center of mass, we have

$\delta {p}_{T} = - l \sin \alpha \delta \alpha \hat{i} + l \cos \alpha \delta \alpha \hat{j}$

Finally

$\delta w = \left\langle\delta {p}_{1} , - m g \hat{j}\right\rangle + \left\langle\delta {p}_{2} , - m g \hat{j}\right\rangle + \left\langle\delta {p}_{T} , - M g \hat{j}\right\rangle = 0$ or

$2 g m r \cos \theta - g l M \cos \alpha = 0$ solving for $M$

$M = \frac{2 m r \cos \theta}{l \cos \left(\alpha\right)}$ but $l \cos \left(\alpha\right) = R$ and $\cos \theta = \frac{R - r}{r}$ so finally

$M = 2 m \left(1 - \frac{r}{R}\right)$

Note:
The detailed calculation of $\delta {p}_{2}$ follows. As we know

${p}_{1} = \left(r - 2 R\right) \hat{i} + r \hat{j}$ and
${p}_{2} = {p}_{1} + 2 r \cos \theta \hat{i} + 2 r \sin \theta \hat{j}$
After a rigid rotation of $\delta \alpha$ counterclockwise the ${p}_{2}$ new coordinates are

$p {'}_{2} = p {'}_{1} + 2 r \cos \left(\theta - \delta \alpha\right) \hat{i} + 2 r \sin \left(\theta - \delta \alpha\right) \hat{j}$

then

$\delta {p}_{2} = p {'}_{2} - {p}_{2}$

but

$\cos \left(\theta - \delta \alpha\right) = \cos \theta \cos \delta \alpha + \sin \theta \sin \delta \alpha$
$\sin \left(\theta - \delta \alpha\right) = \sin \theta \cos \delta \alpha - \cos \theta \sin \delta \alpha$

and also

$\cos \delta \alpha \approx 1$ and $\sin \delta \alpha \approx \delta \alpha$

Finally

$\delta {p}_{2} = r \delta \alpha \hat{i} + 2 r \sin \theta \delta \alpha \hat{i} - 2 r \cos \theta \delta \alpha \hat{j}$

What is the Law of Conservation of Energy??

Pitufox27
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.

Explanation:

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 $h$ on the ground and we are holding an object. When it is released, its initial velocity is zero, but due to the gravitational pull of the Earth, will start moving with a constant acceleration of $9.8 m / {s}^{2}$, so that when it reaches the ground will collide against it at a speed $v$ that if we estimate we will see that is #v = sqrt {2 · g · h} #.

What has happened from the point of view of energy?

When the object was at a height $h$ on the ground with velocity ${v}_{0} = 0$ its gravitational potential energy is #U = m · g · h # and its kinetic energy ${E}_{k} = 0$.

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 $U = 0$. What happened with the potential energy that was the object before falling? If we calculate the kinetic energy just when the object hits the ground, its value is #E_k = 1 // 2 · m cdot v ^ 2 #. If we make the calculations we see that the potential energy loss suffered by the falling body has been transformed into a gain of kinetic energy in the same process.

If $\Delta U = U \left(g r o u n d\right) - U \left(u p\right)$ is the change in potential energy that has taken place in this process, and $\Delta {E}_{k} = {E}_{k} \left(g r o u n d\right) - {E}_{k} \left(u p\right)$ is the corresponding change in energy kinetics, according to the law of conservation of energy has to be that:

$\Delta E = \Delta U + \Delta {E}_{k} = 0 \Rightarrow \Delta U = - \Delta {E}_{k}$ .

Ideal machines,although they do not exist, do not require anything extra. A machine that we actually use can be, in a sense, almost reversible: what message is this line conveying ? SEE THE DETAILS.

Truong-Son N.
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 #100%# efficiency, and the work it performs is #100%# reversible. Real machines, however, can never quite achieve #100%# efficiency, and the work they do is nearly #100%# reversible.

As I derived above, a cyclic process has a change in entropy of $\Delta S = 0$, and if heat flow $q$ is reversible, then $\Delta S = {q}_{\text{rev}} / T$.

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

$\textcolor{b l u e}{e} = | w \frac{|}{q} _ H = \frac{{q}_{H} + {q}_{C}}{q} _ H = \textcolor{b l u e}{1 + {q}_{C} / {q}_{H}}$.

${q}_{C}$ flows out of the engine, and thus, ${q}_{C} \le 0$. When ${q}_{C} = 0$, the efficiency is #100%#.

Or, using the first relation:

$\textcolor{b l u e}{e} = | w \frac{|}{q} _ H = \frac{{q}_{H} + {q}_{C}}{q} _ H = \textcolor{b l u e}{1 - {T}_{C} / {T}_{H}}$.

So, we could even say that since all absolute temperatures are positive (i.e. when in $\text{K}$), as they are in this equation, $0 \le e \le 1$. When ${T}_{C} = {T}_{H}$, $e = 0$, and when ${T}_{C} = \text{0 K}$ (which has yet to be accomplished!), or ${T}_{H}$ is absurdly large, $e \approx 1$.

Therefore, the efficiency can never be more than #100%#, and for real machines, the efficiency is effectively never exactly #100%#.

An astronaut with a mass of #80 kg# is floating in space. If the astronaut throws an object with a mass of #16 kg# at a speed of #3/8 m/s#, how much will his speed change by?

Morgan
Featured 1 week ago

The speed of the astronaut changes by $0.075 \frac{m}{s}$ in the opposite direction of the rock.

Explanation:

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:

${\vec{P}}_{f} = {\vec{P}}_{i}$

For multiple objects,

$\vec{P} = {\vec{p}}_{t o t} = \sum \vec{p} = {\vec{p}}_{1} + {\vec{p}}_{2} + \ldots {\vec{p}}_{n}$

In our case, we have the momentum of the thrown object and the astronaut. We'll call the mass of astronaut ${m}_{1}$ and the mass of the object ${m}_{2}$. So, we're given that ${m}_{1} = 80 k g$, ${v}_{1 i} = 0$, ${m}_{2} = 16 k g$, ${v}_{2 i} = 0 \frac{m}{s}$, and ${v}_{2 f} = \frac{3}{8} \frac{m}{s}$. We want to find ${v}_{1 f}$, the final velocity of the astronaut, and compare this to his original value. We can set up an equation for momentum conservation:

${m}_{1} {v}_{1 i} + {m}_{2} {v}_{2 i} = {m}_{1} {v}_{1 f} + {m}_{2} {v}_{2 f}$

However, both the thrown object and the astronaut are initially at rest, so the total momentum before the "explosion" is $0$.

$0 = {m}_{1} {v}_{1 f} + {m}_{2} {v}_{2 f}$

We can manipulate the equation to solve for ${v}_{1 f}$

${v}_{f 1} = - \frac{{m}_{2} {v}_{f 2}}{m} _ 1$

Using our known values:

${v}_{f 1} = - \frac{\left(16 k g\right) \left(\frac{3}{8} \frac{m}{s}\right)}{80 k g}$

$\implies {v}_{f 1} = - \frac{3}{40} \frac{m}{s} = - 0.075 \frac{m}{s}$

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 $0$, this is also how much his velocity changes by: $- 0.075 \frac{m}{s}$ or $0.075 \frac{m}{s}$ in the opposite direction of the rock.

Hope this helps!

A ball with a mass of #420 g# is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of #28 (kg)/s^2# and was compressed by #9/6 m# when the ball was released. How high will the ball go?

Morgan
Featured 1 week ago

The ball will reach a maximum altitude of $\approx 5.1 m$.

Explanation:

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.

${E}_{m e c h} = K + U$

Where $K$ is kinetic energy and $U$ is potential energy. Thus, when energy is conserved in a system it should follow that

${E}_{i} = {E}_{f}$

${K}_{i} + {U}_{i} = {K}_{f} + {U}_{f}$

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 (${E}_{t h}$). In reality, energy is "lost" to a variety of other sources as well, such as sound.

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.

${U}_{s p} = {U}_{f} + {K}_{f}$

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 ${U}_{g} = m g h$. When the ball reaches its maximum altitude and pauses momentarily before falling, it has no kinetic energy (remember projectile motion?), and therefore we ultimately have only gravitational potential energy.

${U}_{s p} = {U}_{g}$

Spring potential energy, ${U}_{s p}$ is given by #1/2k(Δs)^2#.

#1/2k(Δs)^2=mgh#

( tl;dr ) Rearranging to solve for $h$,

#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 $h$.

$h = \frac{\frac{1}{2} \left(28 \frac{N}{m}\right) {\left(\frac{9}{6} m\right)}^{2}}{0.420 k g \cdot 9.8 \frac{m}{s} ^ 2}$

$h \approx 5.1 m$

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