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## A 1.0 kg object moving at 10 m/s 30° N of E collides centrally with a 2.0 kg object moving at 5 m/s 30° S of W. If the collision is inelastic, what is the velocity of the combined objects after collision?

Morgan
Featured 1 month ago

Inelastic: $0$

Elastic: ${v}_{1 f} = - 10 \text{m"//"s, " v_(2f)=5"m"//"s}$

#### Explanation:

This problem can be solved using (2-D) momentum conservation.

Momentum is conserved in all collisions.

• In an inelastic collision, momentum is conserved as always, but energy is not; part of the kinetic energy is transformed into some other form of energy. For the first situation, we have an inelastic collision.

• The equation for momentum: $\vec{p} = m \vec{v}$

Momentum conservation:

$\Delta \vec{P} = 0$

$\implies \textcolor{\mathrm{da} r k b l u e}{{\vec{P}}_{f} = {\vec{P}}_{i}}$

We are given the following information:

• $\mapsto {m}_{1} = 1.0 \text{kg}$
• $\mapsto {m}_{2} = 2.0 \text{kg}$
• $\mapsto {v}_{1 i} = 10 \text{m"//"s}$
• $\mapsto {v}_{2 i} = 5 \text{m"//"s}$
• $\mapsto {\theta}_{1} = {30}^{o}$
• $\mapsto {\theta}_{2} = {210}^{o}$

Note that I am taking the positive x-axis to be my reference point for all angles.

Because we are given angles for the direction of the velocities of each of the masses, we will need to decompose the vectors into their parallel (x, horizontal) and perpendicular (y, vertical) components. Consequently, we will also have components of momentum.

$\textcolor{\mathrm{da} r k b l u e}{{P}_{\text{net}} = \sqrt{{P}_{x}^{2} + {P}_{y}^{2}}}$

Therefore, by momentum conservation, it should follow that:

${P}_{x} = \sum {p}_{x} = {p}_{x 1} + {p}_{x 2} = {m}_{1} {v}_{1 x} + {m}_{2} {v}_{2 x}$

${P}_{y} = \sum {p}_{y} = {p}_{y 1} + {p}_{y 2} = {m}_{1} {v}_{1 y} + {m}_{2} {v}_{2 y}$

Before and after the collision.

Diagram:

• This shows the paths that the masses would take if they continued in their initial direction.

We can begin by calculating the components of each of the velocities using basic trigonometry. I will demonstrate how this is done for one of the components below.

$\sin \left(\theta\right) = \text{opposite"/"hypotenuse}$

$\implies \sin \left({\theta}_{1}\right) = {v}_{1 y} / {v}_{1}$

$\implies \textcolor{\mathrm{da} r k b l u e}{{v}_{1 y} = {v}_{1} \sin \left({\theta}_{1}\right)}$

Similarly, we find:

• $\textcolor{\mathrm{da} r k b l u e}{{v}_{1 x} = {v}_{1} \cos \left({\theta}_{1}\right)}$
• $\textcolor{\mathrm{da} r k b l u e}{{v}_{2 x} = {v}_{2} \cos \left({\theta}_{2}\right)}$
• $\textcolor{\mathrm{da} r k b l u e}{{v}_{2 y} = {v}_{2} \sin \left({\theta}_{2}\right)}$

We can now calculate the momentum before the collision.

$\textcolor{\mathrm{da} r k b l u e}{{P}_{x} = {m}_{1} {v}_{1 x} + {m}_{2} {v}_{2 x}}$

$\implies = {m}_{1} {v}_{1} \cos \left({\theta}_{1}\right) + {m}_{2} {v}_{2} \cos \left({\theta}_{2}\right)$

$\implies = \left(1.0 \text{kg")(10"m"//"s")cos(30^o)+(2.0"kg")(5"m"//"s}\right) \cos \left({210}^{o}\right)$

$\implies \textcolor{\mathrm{da} r k b l u e}{0}$

$\textcolor{\mathrm{da} r k b l u e}{{P}_{y} = {m}_{1} {v}_{1 y} + {m}_{2} {v}_{2 y}}$

$\implies = {m}_{1} {v}_{1} \sin \left({\theta}_{1}\right) + {m}_{2} {v}_{2} \sin \left({\theta}_{2}\right)$

$\implies = \left(1.0 \text{kg")(10"m"//"s")sin(30^o)+(2.0"kg")(5"m"//"s}\right) \sin \left({210}^{o}\right)$

$\implies \textcolor{\mathrm{da} r k b l u e}{0}$

$\implies \textcolor{\mathrm{da} r k b l u e}{{P}_{\text{net}} = \sqrt{0} = 0}$

By momentum conservation, the momentum after the collision must also be $0$.

Therefore, after the collision, we have:

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

Because it is implied that the masses collide and stick together, they should have a common final velocity.

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

Submitting in our known masses, which do not change:

$\implies {v}_{f} + 2 {v}_{f} = 0$

$\implies 3 {v}_{f} = 0$

$\therefore {v}_{f} = 0$

The final velocity of the masses is zero.

Note that this is, in fact, what is called a perfectly inelastic collision because the objects have a common final velocity.

$- - - - - - -$

In an elastic collision, energy and momentum are both conserved, and the following equations can be derived using a combination of the two:

$\textcolor{\mathrm{da} r k b l u e}{{v}_{1 f} = \left(\frac{{m}_{1} - {m}_{2}}{{m}_{1} + {m}_{2}}\right) {v}_{1 i} + \left(\frac{2 {m}_{2}}{{m}_{1} + {m}_{2}}\right) {v}_{2 i}}$

$\textcolor{\mathrm{da} r k b l u e}{{v}_{2 f} = \left(\frac{2 {m}_{1}}{{m}_{1} + {m}_{2}}\right) {v}_{1 i} + \left(\frac{{m}_{2} - {m}_{1}}{{m}_{1} + {m}_{2}}\right) {v}_{2 i}}$

Using our known values, the parallel components of the final velocities are:

v_(1fx)=((1.0"kg"-2.0"kg")/(1.0"kg"+2.0"kg"))(10cos(30^o)"m"//"s")+((2*2.0"kg")/(1.0"kg"+2.0"kg"))(5cos(210^o)"m"//"s")

$\implies {v}_{1 f x} = \left(- \frac{5 \sqrt{3}}{3} - \frac{20 \sqrt{3}}{9}\right) \text{m"//"s}$

$\implies {v}_{1 f x} = - 5 \sqrt{3} \text{m"//s}$

$\implies \textcolor{\mathrm{da} r k b l u e}{\approx - 8.66 \text{m"//"s}}$

Similarly, $\textcolor{\mathrm{da} r k b l u e}{{v}_{2 f} \approx 4.33 \text{m"//"s}}$

The perpendicular components of the final velocities are:

v_(1fy)=((1.0"kg"-2.0"kg")/(1.0"kg"+2.0"kg"))(10sin(30^o)"m"//"s")+((2*2.0"kg")/(1.0"kg"+2.0"kg"))(5sin(210^o)"m"//"s")

$\implies {v}_{1 f y} = \left(- \frac{10}{6} + \frac{10}{3}\right) \text{m"//"s}$

$\implies \textcolor{\mathrm{da} r k b l u e}{{v}_{1 f y} = - 5 \text{m"//"s}}$

Similarly, $\textcolor{\mathrm{da} r k b l u e}{{v}_{2 f} = 2.5 \text{m"//"s}}$

We will now need to find the magnitude of each final velocity from the components.

${v}_{1 f} = \sqrt{{v}_{1 f x}^{2} + {v}_{1 f y}^{2}}$

$\implies = \sqrt{{\left(8.66\right)}^{2} + {\left(- 5\right)}^{2}}$

$\implies \approx 10 \text{m"/"s}$

Similarly, ${v}_{2 f} \approx 5 \text{m"/"s}$

Since this is an elastic collision and we expect the balls to move off in the opposite direction that they approached each other from, we expect a negative velocity for ${\text{m}}_{1}$ and a positive velocity for ${\text{m}}_{2}$.

Therefore:

• ${v}_{1 f} = - 10 \text{m"//"s}$

• ${v}_{2 f} = 5 \text{m"//"s}$

You can check the answer using momentum conservation. As calculated above, momentum before is $0$. Therefore, momentum after must also be $0$.

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

$\implies \left(1.0 \text{kg")(-10"m"//"s")+(2.0"kg")(5.0"m"//"s}\right)$

$\implies = 0$

## A ball with a mass of 720 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 36 (kg)/s^2 and was compressed by 5/8 m when the ball was released. How high will the ball go?

Morgan
Featured 4 days ago

Maximum altitude $\approx 1.0 \text{m}$

#### Explanation:

$\implies$I have provided a full explanation. If you feel that you have a strong grasp of the concepts but trouble with deriving the final equation, skip to *tl;dr*.

Ignoring any outside forces (air resistance, other sources of friction, etc.) this problem can be solved with energy conservation:

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

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

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

$\implies \textcolor{\mathrm{da} r k b l u e}{{K}_{i} + {U}_{i} = {K}_{f} + {U}_{f}}$

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.

• Initially, assuming the spring is compressed flat against the surface below, 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.

$\textcolor{\mathrm{da} r k b l u e}{{U}_{s p} = {U}_{f} + {K}_{f}}$

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.

• As the ball gains altitude, it gains 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.

Gravitational potential energy:

$\textcolor{\mathrm{da} r k b l u e}{{U}_{g} = m g h}$

• When the ball reaches its maximum altitude and pauses momentarily before falling, it has no kinetic energy, and therefore we ultimately have only gravitational potential energy. Our statement of energy conservation becomes:

$\textcolor{\mathrm{da} r k b l u e}{{U}_{s p} = {U}_{g}}$

Where color(darkblue)(U_(sp)=1/2k(Δs)^2)

$k$ is the spring constant of the spring and $\Delta s$ is its displacement from equilibrium (length).

=>color(darkblue)(1/2k(Δs)^2=mgh)

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

=>color(crimson)(h=(1/2k(Δs)^2)/(mg))

We are given the following information:

• $\mapsto m = 720 \text{g"=0.720"kg}$
• $\mapsto k = 36 {\text{kg"//"s}}^{2}$
• $\mapsto \Delta s = \frac{5}{8} \text{m}$

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

h=(1/2(36"kg"//"s"^2)(5/8"m")^2)/(0.720"kg"*9.81"m"//"s"^2)

$\textcolor{c r i m s o n}{\approx 1.0 \text{m}}$

## Resultant of two equal forces may have the magnitude equal to one of the forces. At what angle between the two equal forces this is possible? Justify your answer.

Nathan L.
Featured 2 weeks ago

$\theta = {120}^{\text{o}}$

#### Explanation:

We're asked to find the necessary angle between two force vectors so that the resultant vector has a magnitude equal to that of one of the forces.

Let's make the two force vectors have magnitudes $A$ and $B$, and their resultant have magnitude $C$. We can use the law of cosines:

${C}^{2} = {A}^{2} + {B}^{2} - 2 A B \cos \theta$

Since the two vector magnitudes $A$ and $B$ are equal, we can just replace $B$ with $A$:

${C}^{2} = 2 {A}^{2} - 2 {A}^{2} \cos \theta$

Now, let's find the angle $\theta$ between the constituent forces:

$\cos \theta = \frac{2 {A}^{2} - {C}^{2}}{2 {A}^{2}}$

And since the resultant force's magnitude is equal to that of one of the vector magnitudes, $C = A$:

$\cos \theta = \frac{2 {A}^{2} - {A}^{2}}{2 {A}^{2}}$

$\cos \theta = \frac{1}{2}$

color(red)(ulbar(|stackrel(" ")(" "theta = 120^"o"" ")|)

The angle between the two force vectors must be color(red)(120^"o".

## Derive expression of the equation of electric potential?

Aritra G.
Featured 1 week ago

Consider a point charge $q$.

We are free to chose the origin anywhere we please. In this case, we do it at the location of the charge.

Thus, the field due to the charge at a distance $r$ from the charge is by Coulomb's law,

$\vec{E} = \frac{1}{4 \pi {\epsilon}_{0}} \frac{q}{r} ^ 2 \hat{r}$ where $\hat{r}$ is the radially outward unit vector.

But, we know that electric potential $V$ is negative gradient of electric field,

$\vec{E} = - \nabla V$

Thus, employing spherical polar coordinates,

$\vec{E} = - \frac{\partial V}{\partial r} \hat{r}$

implies 1/(4piepsilon_0)q/r^2hatr = -(delV)/(delr)hatr

Integrating,

$V = - {\int}_{r}^{\propto} \frac{1}{4 \pi {\epsilon}_{0}} \frac{q}{r} ^ 2 \mathrm{dr}$

$V = {\int}_{\propto}^{r} \frac{1}{4 \pi {\epsilon}_{0}} \frac{q}{r} ^ 2 \mathrm{dr}$

$\implies V = \frac{q}{4 \pi {\epsilon}_{0}} \left[\frac{1}{r} - \frac{1}{\propto}\right]$

But, $\frac{1}{\propto} = 0$

$\implies V = \frac{q}{4 \pi {\epsilon}_{0} r}$

This is the expression for electrostaic potential due to a point charge $q$ at a distance $r$ from it.

## Electric field?

A08
Featured 1 week ago

I get
(A) and (C)

#### Explanation:

Given electric field intensity at a point $\left(x , y\right)$ as $\vec{I} = \left(12 x {y}^{3} - 4 x\right) \hat{i} + 18 {x}^{2} {y}^{2} \hat{j}$

Let's examine the case of a two-dimensional vector field whose
$\text{*}$scalar curl is =∂/(∂x)F_2−∂/(∂y)F_1

For the given field we have

∂/(∂x)(18x^2y^2)−∂/(∂y)(12xy^3-4x)
=36xy^2−(36xy^2-0)
$= 0$

As scalar curl $= 0$, the function represents a conservative electric field.

We know that potential in an two dimensional electric field is expressed as

$\vec{E} = - \left[\hat{i} \frac{\partial}{\partial x} + \hat{j} \frac{\partial}{\partial y}\right] V \left(x , y\right)$

For the given electric field above

$- \frac{\partial f}{\partial x} = \left(12 x {y}^{3} - 4 x\right)$ ....(1) and $- \frac{\partial f}{\partial y} = 18 {x}^{2} {y}^{2}$ .....(2)

From equation (1) using partial integration

$- f \left(x , y\right) = \int \left(12 x {y}^{3} - 4 x\right) \mathrm{dx}$
$\implies - f \left(x , y\right) = \left(12 \times {x}^{2} / 2 {y}^{3} - 4 \times {x}^{2} / 2 + C \left(y\right)\right)$
where $C \left(y\right)$ is a constant of integration dependent on $y$.
$\implies - f \left(x , y\right) = \left(6 {x}^{2} {y}^{3} - 2 {x}^{2} + C \left(y\right)\right)$

Differentiating this with respect to $y$ and setting equal to equation (2) we get

$- \frac{\partial f \left(x , y\right)}{\partial y} = - \frac{\partial}{\partial y} \left(6 {x}^{2} {y}^{3} + 2 {x}^{2} + C \left(y\right)\right) = - 18 {x}^{2} {y}^{2}$
$\implies \left(6 {x}^{2} \left(3 {y}^{2}\right) + \frac{d}{\mathrm{dy}} C \left(y\right)\right) = 18 {x}^{2} {y}^{2}$
$\implies \frac{d}{\mathrm{dy}} C \left(y\right) = 0$
$\implies C \left(y\right)$ is actually a constant independent of both $x \mathmr{and} y$.

The potential function becomes

$f \left(x , y\right) = \left(6 {x}^{2} {y}^{3} - 2 {x}^{2} + c\right)$

Given that at origin electric potential is zero. Therefore $c = 0$ and potential function reduces to

$f \left(x , y\right) = 6 {x}^{2} {y}^{3} - 2 {x}^{2}$ ......(3)

Therefore, electric potential at point $\left(1 , 1\right)$ is found from equation (3) as

$- f \left(1 , 1\right) = - 4 V$

.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-

A two dimensional vector field $\vec{F}$ given as

$\vec{F} = {F}_{1} \hat{i} + {F}_{2} \hat{j}$

is conservative if partial derivative

∂/(∂x)F_2−∂/(∂y)F_1=0

The LHS is also called $\text{*}$scalar curl.
......................................

To determine whether a three dimensional vector field is conservative we find a function $f$ such that $\vec{F} = \nabla f$.
Therefore if curl $\vec{I} = 0$, then $\vec{I}$ is conservative.

Where curl $\vec{I}$ is given as
$\nabla \times \vec{I} = \left[\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ \hat{i} \text{ part" & hatj" part" & hatk" part}\end{matrix}\right]$

## A metal sphere falls with terminal velocity through cylinder filled with oil.If sphere experiences a viscous force of kv(k is constant), what is the kinetic energy of the sphere(express in m,g,k form)?

A08
Featured yesterday

This is what I get

#### Explanation:

When metal sphere is moving at terminal velocity in a cylinder filled with oil, the forces acting on it are in equilibrium.

$\text{Frictional force" uarr="weight" darr-"upthrust} \uparrow$ .....(1)

Given frictional force $= k {v}_{0}$.
Hence above equation becomes

$k {v}_{0} = \left({m}_{s} - {m}_{o}\right) g$
where ${v}_{0}$ is the terminal velocity, ${m}_{s}$ is mass of metal sphere and ${m}_{o}$ is mass of oil displaced.
$\implies {v}_{0} = \left({m}_{s} - {m}_{o}\right) \frac{g}{k}$ ......(2)

Kinetic energy of the sphere is given as

$K E = \frac{1}{2} {m}_{s} {v}_{0}^{2}$

Inserting value of terminal velocity from (2) we get

$K E = \frac{1}{2} {m}_{s} {\left(\left({m}_{s} - {m}_{o}\right) \frac{g}{k}\right)}^{2}$
$K E = \frac{1}{2} {m}_{s} {g}^{2} / {k}^{2} {\left({m}_{s} - {m}_{o}\right)}^{2}$
.-.-.-.-.-.-.-.-.-.-.-

Using Stokes Law for the frictional force (of viscosity) on a small sphere moving through a viscous fluid for LHS of (1) we get

$6 \pi \eta \alpha {v}_{0} = \frac{4}{3} \pi {\alpha}^{3} g \left({\rho}_{s} - {\rho}_{o}\right)$ ......(1)
where η is viscosity of oil, $\alpha$ is radius of the metal sphere, ${v}_{0}$ is terminal velocity, $g$ is acceleration due to gravity, ρ_s is density of the material of sphere and ${\rho}_{o}$ is density of oil.
$\implies k = 6 \pi \eta \alpha$, ${m}_{s} = \frac{4}{3} \pi {\alpha}^{3} {\rho}_{s} \mathmr{and} {m}_{o} = \frac{4}{3} \pi {\alpha}^{3} {\rho}_{o}$ in the given question.

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