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Inelastic:
Elastic:
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:
Momentum conservation:
#DeltavecP=0#
#=>color(darkblue)(vecP_f=vecP_i)#
We are given the following information:
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.
#color(darkblue)(P_"net"=sqrt(P_x^2+P_y^2))#
Therefore, by momentum conservation, it should follow that:
#P_x=sump_x=p_(x1)+p_(x2)=m_1v_(1x)+m_2v_(2x)#
#P_y=sump_y=p_(y1)+p_(y2)=m_1v_(1y)+m_2v_(2y)#
Before and after the collision.
Diagram:
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(theta)="opposite"/"hypotenuse"#
#=>sin(theta_1)=v_(1y)/v_1#
#=>color(darkblue)(v_(1y)=v_1sin(theta_1))#
Similarly, we find:
We can now calculate the momentum before the collision.
#color(darkblue)(P_x=m_1v_(1x)+m_2v_(2x))#
#=>=m_1v_1cos(theta_1)+m_2v_2cos(theta_2)#
#=>=(1.0"kg")(10"m"//"s")cos(30^o)+(2.0"kg")(5"m"//"s")cos(210^o)#
#=>color(darkblue)(0)#
#color(darkblue)(P_y=m_1v_(1y)+m_2v_(2y))#
#=>=m_1v_1sin(theta_1)+m_2v_2sin(theta_2)#
#=>=(1.0"kg")(10"m"//"s")sin(30^o)+(2.0"kg")(5"m"//"s")sin(210^o)#
#=>color(darkblue)(0)#
#=>color(darkblue)(P_"net"=sqrt(0)=0)#
By momentum conservation, the momentum after the collision must also be
Therefore, after the collision, we have:
#P_f=m_1v_(f1)+m_2v_(f2)=0#
Because it is implied that the masses collide and stick together, they should have a common final velocity.
#=>m_1v_f+m_2v_f=0#
Submitting in our known masses, which do not change:
#=>v_f+2v_f=0#
#=>3v_f=0#
#:.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:
#color(darkblue)(v_(1f)=((m_1-m_2)/(m_1+m_2))v_(1i)+((2m_2)/(m_1+m_2))v_(2i))#
#color(darkblue)(v_(2f)=((2m_1)/(m_1+m_2))v_(1i)+((m_2-m_1)/(m_1+m_2))v_(2i))#
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")#
#=>v_(1fx)=(-(5sqrt3)/3-(20sqrt3)/9)"m"//"s"#
#=>v_(1fx)=-5sqrt3"m"//s"#
#=>color(darkblue)(~~-8.66"m"//"s")#
Similarly,
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")#
#=>v_(1fy)=(-10/6+10/3)"m"//"s"#
#=>color(darkblue)(v_(1fy)=-5"m"//"s")#
Similarly,
We will now need to find the magnitude of each final velocity from the components.
#v_(1f)=sqrt(v_(1fx)^2+v_(1fy)^2)#
#=>=sqrt((8.66)^2+(-5)^2)#
#=>~~10"m"/"s"# Similarly,
#v_(2f)~~5"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
Therefore:
You can check the answer using momentum conservation. As calculated above, momentum before is
#P_f=m_1v_(1f)+m_2v_(2f)#
#=>(1.0"kg")(-10"m"//"s")+(2.0"kg")(5.0"m"//"s")#
#=>=0#
Maximum altitude
Ignoring any outside forces (air resistance, other sources of friction, etc.) this problem can be solved with energy conservation:
#DeltaE_(mech)=DeltaK+DeltaU#
Where
#E_i=E_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.
#color(darkblue)(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.
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:
#color(darkblue)(U_g=mgh)#
#color(darkblue)(U_(sp)=U_g)#
Where
#k# is the spring constant of the spring and#Deltas# is its displacement from equilibrium (length).
#=>color(darkblue)(1/2k(Δs)^2=mgh)#
( tl;dr ) Rearranging to solve for
We are given the following information:
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(36"kg"//"s"^2)(5/8"m")^2)/(0.720"kg"*9.81"m"//"s"^2)#
#color(crimson)(~~1.0"m")#
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
#C^2 = A^2 + B^2 - 2ABcostheta#
Since the two vector magnitudes
#C^2 = 2A^2 - 2A^2costheta#
Now, let's find the angle
#costheta = (2A^2 - C^2)/(2A^2)#
And since the resultant force's magnitude is equal to that of one of the vector magnitudes,
#costheta = (2A^2 - A^2)/(2A^2)#
#costheta = 1/2#
#color(red)(ulbar(|stackrel(" ")(" "theta = 120^"o"" ")|)#
The angle between the two force vectors must be
Consider a point charge
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
But, we know that electric potential
Thus, employing spherical polar coordinates,
Integrating,
But,
This is the expression for electrostaic potential due to a point charge
I get
(A) and (C)
Given electric field intensity at a point
Let's examine the case of a two-dimensional vector field whose
For the given field we have
As scalar curl
We know that potential in an two dimensional electric field is expressed as
#vecE=-[hatidel/(delx)+hatjdel/(dely)]V(x,y)#
For the given electric field above
#-(delf)/(delx)=(12xy^3-4x)# ....(1) and#-(delf)/(dely)=18x^2y^2# .....(2)
From equation (1) using partial integration
#-f(x,y)=int(12xy^3-4x)dx#
#=>-f(x,y)=(12xxx^2/2y^3-4xxx^2/2+C(y))#
where#C(y)# is a constant of integration dependent on#y# .
#=>-f(x,y)=(6x^2y^3-2x^2+C(y))#
Differentiating this with respect to
#-(delf(x,y))/(dely)=-del/(dely)(6x^2y^3+2x^2+C(y))=-18x^2y^2#
#=>(6x^2 (3y^2)+d/dyC(y))=18x^2y^2#
#=>d/dyC(y)=0#
#=>C(y)# is actually a constant independent of both#x and y# .
The potential function becomes
#f(x,y)=(6x^2y^3-2x^2+c)#
Given that at origin electric potential is zero. Therefore
#f(x,y)=6x^2y^3-2x^2# ......(3)
Therefore, electric potential at point
#-f(1,1)=-4V#
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
A two dimensional vector field
#vecF=F_1hati+F_2hatj#
is conservative if partial derivative
#∂/(∂x)F_2−∂/(∂y)F_1=0#
The LHS is also called
......................................
To determine whether a three dimensional vector field is conservative we find a function
Therefore if curl
Where curl
This is what I get
When metal sphere is moving at terminal velocity in a cylinder filled with oil, the forces acting on it are in equilibrium.
#"Frictional force" uarr="weight" darr-"upthrust" uarr# .....(1)
Given frictional force
Hence above equation becomes
#kv_0=(m_s-m_o)g#
where#v_0# is the terminal velocity,#m_s# is mass of metal sphere and#m_o# is mass of oil displaced.
#=>v_0=(m_s-m_o)g/k# ......(2)
Kinetic energy of the sphere is given as
#KE=1/2m_sv_0^2#
Inserting value of terminal velocity from (2) we get
#KE=1/2m_s((m_s-m_o)g/k)^2#
#KE=1/2m_sg^2/k^2(m_s-m_o)^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
#6pietaalphav_0=4/3pialpha^3g(rho_s-rho_o)# ......(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.
#=>k=6pietaalpha# ,#m_s=4/3pialpha^3rho_s and m_o=4/3pialpha^3rho_o# in the given question.
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