Problem Based Learning in Basic Physics - X

Problem Based Learning in Basic Physics - X

 

                                                       A. K .Mody

C-14 Beverly Hills, Lam Road, Devlali, Nashik 422401, Maharashtra

                                                                       H. C. Pradhan

HBCSE, TIFR, V. N. Purav Marg, Mankhurd

Mumbai – 400 088

[In this article- tenth in the series of articles we present problems for a problem based learning course based on system with continuously varying parameter.]

In this article 10^th in the PBL series we consider various examples from different areas of physics where we have systems in which one of the parameter varies continuously and to estimate any effect we need to take differential element and perform integration separately (diefferent from that which is already performed in arriving at standard text book formula).

1.      A block of mass m sliding on a frictionless horizontal surface experiences an air resistance proportional to its velocity. If it has initial speed v₀, find an expression for its velocity as a function of time.

Solution:

Acceleration

                        

Integrating,  

2.      A car starts from rest and start moving while its engine continues to deliver constant power P. Find the distance travelled by the car in time t.

Solution:

In this case, since constant power is delivered, acceleration is not constant.

Power delivered                           

;     integrating which, we get

;     integrating which, we get

3.      A cylindrical solid of mass 10 ^{– 2} kg and cross-sectional area 10 ^{– 4} m² is moving parallel to its axis (the x-axis) with a uniform speed of 10³ m/s in the positive direction. At t = 0, its front face passes the plane x = 0. The region to the right of this plane is filled with dust particle of uniform density 10 ^{– 3} kg/m³. When a dust particle collides with the face of the cylinder, it sticks to its surface. Assuming that the dimension of the cylinder practically remains unchanged and the dust sticks only to the front face of the x-coordinate of the front of the cylinder. Find the x-coordinate of the front of the cylinder at t = 150 s.                 [JEE 1993]

Solution:

In this case we have an object (cylinder) whose mass keeps on increasing (variable mass problem) as  m = m₀ +ρAx  and equation will be   .  If we assume mass of the cylinder as m₀ and velovity at t = 0 as v₀, then integration yields .

The same would also follow from conservation of total momentum.

Further integration yields,  .

Upon substitution of parameters gives x = 10⁵ m   at t = 150 s.

4.      A uniform chain of length L is kept on a smooth frictionless horizontal table with small length hanging from the edge. This makes chain slide down from the table. Find the speed when it just loses contact with the table.

Solution:

Let initially small length hanging from the table. Thus gravity pulling it down with the acceleration         

                       

                       

                       

Integrating we get  

When chain just leaves the table, its vertical length x →L

                           is the speed when it just loses contact with the table.

5.      Calculate the gravitational potential energy of a uniform spherical distribution of mass M and radius R.

Solution:

When the a mass dm comes from infinity to the surface of sphere of mass m and radius r, it potential energy is

When mass  accumulates and grows from zero to M and radius from zero to R, potential energy becomes

Since mass distribution is uniform, density

This gives   and integral becomes

Which upon integration gives

[Calculation of electrostatic potential energy of a charged sphere is  identical to the one discussed here for gravitational potential energy.]

6.      For a particle performing SHM with time period T, calculate time required to travel from a/2 to a, where a is the amplitude.

Solution:

For a particle performing SHM,      velocity         where x is displacement                         and ,

Thus,      which gives 

 

7.      A uniform rope of mass 0.4 kg and length 2.45 m hangs from a ceiling.

(i)     Find an expression for increase in length of the rope.

(ii)   Find the speed of the transverse wave in the rope at a point 0.5 m distance from the lower end.

(iii)  Calculate the time taken by a transverse wave to travel the full length of the rope. Derive the formula used. [Take g = 9.8 m/s², and Y = 2×10¹¹N/m² ].  [(ii) and (iii) Roorkee 1991]

Solution:

Consider an element of length dx at a distance x from the lower end of the rope.

The stress experienced by this element is given by ρxg where ρ is the density of the rope.

If length this element increases by dΔx, then strain = .  Here Y is the Young’s modulus of the material of the rope.

 increase in length of this element

 increase in length of the rope  

where  M is the mass of the rope and A is its area of cross-section.

(ii)  velocity of the wave at a distance x from the lower end                    here T is the tension at a distance x from the lower end and m is linear density of the rope.

(iii)  time taken by wave to travel the rope

8.      A string tied between x = 0 and x = l vibrates in fundamental mode. The amplitude A, tension T and mass per unit length µ is given. Find the total energy of the string. [JEE 2003]

Solution:

The amplitude at any point x is given by a = A sinkx where

Where ,  mass of string element of length dx is dm = µdx

Energy of this element while oscillating is

Thus total energy of the string

Which upon integration yields

9.      The capacitance of a parallel plate capacitor with plate are A and separation d is C. The space between two wedges of dielectric constants K₁ and K₂, respectively. Find the capacitance of the resulting capacitor. [JEE 1996]

Solution:

Let A = bL,   where b is the width of the capacitor into the plane of the paper.

Consider a strip of length dx at a distance x from one end as shown below.

The corresponding thicknesses are y and (d – y). Notice here that

Each of these strips have capacitance  and

The series combination gives capacitance of the entire strip dC given by  =

       

The capacitance C of the entire capacitance therefore can be obtained by integrating dC.

10.  A ray of light travelling in air is incident at grazing angle (incident angle 90^o) on a long rectangular slab of a transparent medium of thickness t = 1.0 m (see fig. below). The point of incidence is the origin A(0, 0). The medium has a variable index of refraction µ(y) given by , where k = 1.0 (metre) ^{– 3/2}.

The refractive index of air is 1.0

(A) Obtain the relation between the slope of the trajectory of the ray at a point B(x, y) in the medium and the incident angle at that point.

(B)  Obtain the equation of the trajectory y(x) of the ray in the medium.

(C)  Determine the coordinates (x₁, y₁) of  the point P, where the ray intersects the upper surface of the slab boundary.

(D) Indicate the path of the ray subsequently. [JEE 1995]

Solution:

Note here that refractive index µ depends on co-ordinate y only.

(A)  For a series of interfaces, Snell’s law is µ₁ sinθ₁ = µ₂ sinθ₂ =….= µ_n sinθ_n

Thus at the point of incidence  and point B, the relation is µ₁ sinθ₁ = µ_y sinθ_y and θ₁ ≈ 90^o.

(B) Which (referring to figure gives µ(y) sinθ_y = 1    and slope = dy/dx = tan (90 - θ_y) = cot θ_y 

        which upon integration yields .

(C)  Point P is at y₁  = 1.0 m which gives x₁ = 4.0 m

(D) At point A and P µ_A sinθ_A = µ_P sinθ_P where θ_A ≈ 90^o where as both µ_A = µ_P = 1 as subsequently at P ray enters air. This gives θ_P ≈ 90^o

References:

1. JEE - Joint Entrance Examination for admission to IIT

2.      Roorke : Roorke University Entrance exam (Now IIT-Roorke)