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51.

In a certain two-dimensional field of force the potencial energy of a particle has the uniform `U=alphax^2+betay^2`, where `alpha` and `beta` are positive constants whose magnitudes are different. Find out: (a) whether this field is central, (b) what is the shape of the equipotential surfaces and also of the surfaces for which the magnitude of the vector of force `F=const.`

Answer» (a) We have
`F_x=-(deltaU)/(dx)=-2alphax` and `F_y=(-deltaU)/(dy)=-2betay`
So, `vecF=2alphaxveci-2betaveci` and, `F=2sqrt(alpha^2x^2+beta^2y^2)` (1)
For a central force, `vecrxxvecF=0`
Hence, `vecrxxvecF=(xveci+yvecj)xx(-2alphaxveci-2betayvecj)`
`=-2betaxyveck-2alphaxy(veck)~~0`
Hence the force is not a central force.
(b) As `U=alphax^2+betay^2`
So, `F_x=(deltaU)/(deltax)=-2alphax` and `F_y=(-deltaU)/(deltay)=-2betay`.
So, `F=sqrt(F_x^2+F_y^2)=sqrt(4alpha^2x^2+4beta^2y^2)`
According to the problem
`F=2sqrt(alpha^2x^2+beta^2y^2)=C` (constant)
or, `alpha^2x^2+beta^2y^2=(C^2)/(2)`
or, `(x^2)/(beta^2)+(y^2)/(alpha^2)=(C^2)/(2alpha^2beta^2)=k` (say) (2)
Therefore the surfaces for which F is constant is an ellipse.
For an equipotential surface `U` is constant.
So, `alphax^2+betay^2=C_0` (constant)
or, `(x^2)/(sqrt(beta^2))+(y^2)/(sqrt(alpha^2))=(C_0)/(alphabeta)=K_0` (constant)
Hence teh equipotential surface is also an ellipse.
Discussion
52.

A car moves with a constant tangential acceleration `w_tau=0.62m//s^2` along a horizontal surface circumscribing a circle of wheels of the car and the surface is `k=0.20`. What distance will the car ride without sliding if at the initial moment of time its velocity is equal to zero?

Answer» As initial velocity is zero thus
`v^2=2w_ts` (1)
As `w_tgt0` the speed of the car increases with time or distance. Till the moment, sliding starts, the static friction provides the required centripetal acceleration to the car.
Thus `f r =mw`, but `f r lekmg`
So, `w^2lek^2g^2` or, `w_t^2+v^2/Rlek^2g^2`
or, `v^2le(k^2g^2-w_t^2)R`
Hence `v_(max)=sqrt((k^2g^2-w_t^2)R)`
so, from Eqn. (1), the sought distance `s=(v_(max)^2)/(2w_t)=1/2sqrt(((kg)/(w_t))^2-1)=60m`.
Discussion
53.

Two identical buggies move one after the other due to inertia (without friction) with the same velocity `v_0`. A man of mass m rides the rear buggy. At a certain moment the man jumps into the front buggy with a velocity u relative to his buggy. Knowing that the mass of each buggy is equal to M, find the velocities with which the buggies will move after that.

Answer» From momentum conservation, for the system "rear buggy with man"
`(M+m)vecv_0=m(vecu+vecv_R)+Mvecv_R` (1)
From the momentum conservation, for the system (front buggy+man coming from near buggy)
`Mvecv_0+m(vecu+vecv_R)=(M+m)vecv_F`
So, `vecv_F=(Mvecv_0)/(M+m)+(m)/(M+m)(vecu+vecv_R)`
Putting the value of `vecv_R` from (1), we get
`vecv_F=vecv_0+(mM)/((M+m)^2)vecu`
Discussion
54.

A cyclist rides along the circumference of a circular horizontal plane of radius `R`, with the friction coefficient `mu=mu_(0)(1-(r )/(R ))`, where `mu_(0)` is constant and `r` is distance from centre of plane `O`. Find the radius of the circle along which the cyclist can ride with the maximum velocity, what is this valocity?

Answer» According to the question, the cyclist moves along the circular path and the centripetal force is provided by the frictional force. Thus from the equation `F_n=mw_n`
`f r=(mv^2)/(r)` or `kmg=(mv^2)/(l)`
or `k_0(1-r/R)g=v^2/r` or `v^2=k_0(r-r^2//R)g` (1)
For `v_(max)`, we should have `(d(r-r^2/R))/(dr)=0`
or, `1-(2r)/(R)=0`, so `r=R//2`
Hence `v_(max)=1/2sqrt(k_0gR)`
Discussion
55.

The inclined plane of figure forms an angle `alpha=30^@` with the horizontal. The mass ratio `m_2//m_1=eta=2//3`. The coefficients of friction between the body `m_1` and the inclined plane is equal to `k=0.10`. The masses of the pulley and the threads are negligible. Find the magnitude and the direction of acceleration of the body `m_2` when the formely stationary system of masses starts moving.

Answer» Form the conditions, obtained in the previous problem, first we will check whether the mass `m_2` goes up or down.
Here, `m_2//m_1=etagtsin alpha+kcos alpha`, (substituting the values). Hence the mass `m_2` will come down with an acceleration (say w). From the free body diagram of previous problem,
`m_2-g-T=m_2w` (1)
and `T-m_1gsin alpha-km_1gcosalpha=m_1w` (2)
Adding (1) and (2), we get,
`m_2g-m_1g sin alpha-km_1g cos alpha=(m_1+m_2)w`
`w=((m_2//m_1-sin alpha-kcos alpha)g)/((1+m_2//m_1))=((eta-sin alpha-k cos alpha)g)/(1+eta)`
Substituting all the values, `w=0*048g~~0*05g`
As `m_2` moves down with acceleration of magnitude `w=0*05ggt0`, thus in vector form acceleration of `m_2`:
`vecw_2=((eta-sin alpha-k cos alpha)vecg)/(1+eta)=0.05vecg`.
Discussion
56.

In the system shown in figure the masses of the bodies are known to be `m_1` and `m_2`, the coefficient of friction between the body `m_1` and the horizontal plane is equal to `k`, and a pulley of mass m is assumed to be a uniform disc. The thread does not slip over the pulley. At the moment `t=0` the body `m_2` starts descending. Assuming the mass of the thread and the friction in the axle of the pulley to be negligible, find the work performed by the friction forces acting on the body `m_1` over the first t seconds after the beginning of motion.

Answer» As the system (`m+m_1+m_2`) is under constant forces, the acceleration of body `m_1` an `m_2` is constant. In addition to it the velocities and acceleration of bodies `m_1` and `m_2` at equal in magnitude (say v and w) because the length of the thread is constant.
From the equation of increament of mechanical energy i.e. `DeltaT+DeltaU=A_(f r)`, at time t whe block `m_1` is distance h below from initial position corresponding to `t=0`,
`1/2(m_1+m_2)v^2+1/2((mR^2)/(2))v^2/R^2-m_2gh=-km_1gh` (1)
(as angular velocity `omega=v//R` for no slipping of thread).
But `v^2=2wh`
So using it in (1), we get
`w=(2(m_2-km_1)g)/(m+2(m_1+m_2))` (2)
Thus the work done by the friction force on `m_1`
`A_(f r)=-km_1gh=-km_1g(1/2wt^2)`
`=-(km_1(m_1-km_1)g^2t^2)/(m+2(m_1+m_2))` (using 2)
Discussion
57.

Two small discs of masses `m_1` and `m_2` interconnected by a weightless spring rest on a smooth horizontal plane. The discs are set in motion with initial velocities `v_1` and `v_2` whose directions are mutually perpendicular and lie in a horizontal plane. Find the total energy `overset~E` of this system in the frame of the centre of inertia.

Answer» As initially `U=overset~U=0`, so, `overset~E=overset~T`
From the solution of
`overset~T=1/2mu|vecv_1-vecv_2|`,
As `vecv_1_|_vecv_2`
Thus `overset~=1/2(m_1m_2)/(m_1+m_2)(v_1^2+v_2^2)`
Discussion
58.

A particle of mass `1.0g` moving with velocity `v_1=3.0i-2.0j` experiences a perfectly inelastic collision with another particle of mass `2.0g` and velocity `v_2=4.0j-6.0k`. Find the velocity of the formed particle (both the vector v and its modulus), if the components of the vectors `v_1` and `v_2` are given in the SI units.

Answer» From conservation of momentum, for the closed system "both colliding particles"
`m_1vecv_1+m_2vecv_2=(m_1+m_2)vecv`
or, `vecv=(m_1vecv_1+m_2vecv_2)/(m_1+m_2)=(1(3veci-2vecj)+2(4vecj-6veck))/(3)=veci+2vecj-4veck`
Hence `|vecv|=sqrt(1+4+16)m//s=4*6m//s`
Discussion
59.

The proper lifetime of an unstable particle is equal to `Deltat_0=10ns`. Find the distance this particle will traverse till its decay in the laboratory frame of reference, where its lifetime is equal to `Deltat=20ns`.

Answer» The distance travelled in the laboratory frame of reference is `vDeltat` where v is the velocity of the particle. But by time dilation
`Deltat=(Deltat_0)/(sqrt(1-v^2//c^2)` So `v=csqrt(1-(Deltat_0//Deltat)^2)`
Thus the distance traversed is
`cDeltatsqrt(1-(Deltat_0//Deltat)^2)`
Discussion
60.

A rocket ejects a steady jet whose velocity is equal to u relative to the rocket. The gas discharge rate equals `mu kg//s`. Demonstrate that the rocket motion equation in this case takes the form `mw=F-muu`, where m is the mass of the rocket at a given moment, w is its acceleration, and F is the external force.

Answer» Suppose that at time `t`, the rocket has the mass m and the velocity `vecv`, relative to the reference frame, employed. Now consider the inertial frame moving with the velocity that the rocket has at the given moment. In this reference frame, the momentum increament that the rocket & ejected gas system acquires during time `dt` is,
`dvecp=mdvecv+mudtvecu=vecFdt`
or, `m(dvecv)/(dt)=vecF-muvecu`
or, `mvecw=vecF-muvecu`
Discussion
61.

A relativistic rocket emits a gas jet with non-relativistic velocity u constant relative to the rocket. Find how the velocity v of the rocket depends on its rest mass m if the initial rest mass of the rocket equals to `m_0`.

Answer» The velocity of ejected gases is u relative to the rocket. In an earth centred frame it is
`(v-u)/(1-(vu)/(c^2))`
in the direction of the rocket. The momentum conservation equation then reads
`(m+dm)(v+dv)+(v-u)/(1-(uv)/(c^2))(-dm)=mv`
or `mdv-((v-u)/(1-(uv)/(c^2))-v)dm=0`
Here `-dm` is the mass of the ejected gases. so
`mdv-(-u+(uv^2)/(c^2))/(1-(uv)/(c^2))dm=0`, or `mdv+u(1-v^2/c^2)dm=0`
(neglecting `1-(uv)/(c^2)` since u is non-relativistic.)
Integrating `(beta=v/c), int (dbeta)/(1-beta^2)+u/cint(dm)/(m)=0, 1n(1+beta)/(1-beta)+u/c1nm=const ant`
The constant `=u/c1nm_0` since `beta=0` initially.
Thus `(1-beta)/(1+beta)=(m/m_0)^(u//c)` or `beta=(1-(m/m_0)^(u//c))/(1+(m/m_0)^(u//c))`
Discussion
62.

Proceeding from the fundamental equation of relativistic dynamics, find: (a) under what circumstances the acceleration of a particle coincides in direction with the force F acting on ti, (b) the proportionality factors relating the force F and the acceleration w in the cases when `F_|_v` and `F||v`, where v is the velocity of the particle.

Answer» `vecF=(d)/(dt)((m_0vecv)/(sqrt(1-v^2/c^2)))=m_0(vecv)/(sqrt(1-v^2/c^2))+m_0vecv/c^2vecv*vecv(1)/((1-v^2/c^2)^(3//2))`
Thus `vecF_(_|_)=m_0(vecw)/(sqrt(1-beta^2)), vecw=vecv, vecw_|_vecv`
`vecF_(||)=m_0(vecw)/((1-beta^2)^(3//2)), vecw=vecv, vecw_(||)vecv`
Discussion