Problem Based Learning in Basic Physics - II

Problem Based Learning in Basic Physics - II

 School Science 49 (4) Dec 2011

A.  K . Mody

V. E.S. College of Arts, Science and Commerce, Sindhi Society

Chembur, Mumbai – 400 071

H. C. Pradhan

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

Mumbai – 400 088

In this article- second in the series of articles we present problems for a problem based learning course from the area of mechanics based on Newton’s laws, conservation principles, static equilibrium and rotational motion. We present the learning objectives in this area of basic physics and what each problem tries to achieve with its solution.

 

In this article, second in the series of Problem Based Learning in Basic Physics, we present a problem on collision which is based on conservation laws accompanied by auxiliary problems and problems based on Newton’s laws of motion, static equilibrium and rotational motion.  Methodology and philosophy of selecting these problems are already discussed. (Pradhan 2009, Mody 2011) This article deals with problems where cause of motion has to be understood to be able to predict the motion which is in terms of conservation laws and Newton’s Laws of motion for force as well as torque.

To review methodology in brief,  we note here that this PBL (Problem Based Learning) starts after students have been introduced to formal structure of Physics. Ideally students would attempt only main problem. If they find it difficult,  then depending upon their area of difficulty, right auxiliary problem have to be introduced by teacher who is expected to be a constructivist facilitator. The auxiliary problems suggested here, are only a suggestion. Teacher may choose as per his/her requirement or may construct questions on the spot to guide student to right idea and method.

B: Problems on Collision

4. Collision

A neutron of kinetic energy 65 eV collides inelastically with a singly ionized helium atom at rest. It is scattered at an angle of 90⁰ with respect of its original direction. (i) Find the allowed values of the energy of the neutron and that of the atom after collision.  (ii) If the atom gets de-excited subsequently by emitting radiation, find the frequencies of the emitted radiation.  [Given: mass of He atom = 4´ mass of neutron, ionization energy of H-atom = 13.6eV][JEE 1993]

• This problem creates a scene where students are expected to think about a collision process at an atomic level and subsequent processes involving emission of radiation.

Tasks involved in this problem are:

1. To understand it as an inelastic collision.
2. To understand how energy and momentum are exchanged in such a collision.
3. To apply Conservation of momentum to an inelastic process
4. To realize that Conservation of energy can be applied to inelastic processes if it can be accounted for.
5. To realize that energy levels of He⁺ are identical to H-atom.
6. To see all possible paths of returning to ground state.

The following are the smaller problems used.

A 4.1^*  A paricle of mass m undergoes an elastic collission with a stationary particle of mass M. What is the fraction of kinetic energy lost by the striking particle if

a)      it moves at right angles to its original direction after the collission,

b)      the collission is head-on one?

      This problem allows students to work through one as well as two-dimensional elastic collisions and there by become familiar with some of the tasks involved. Here students work with conservation of momentum and kinetic energy, as this is elastic collision.

A 4.2  A bullet of mass 0.012 kg and horizontal speed  70 m/s strikes a block of a wood of mass 0.4 kg, and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block. [NCERT 2006]

     This problem allows students to work through a one dimensional collision and realsie what may happen to energy if the collision is inelastic. At the same time it needs application of conservation total energy as block plus bullet system moves in gravitational field.

1. In addition students have to be guided to understand inelastic process at an atomic level through relevant questions.
2. Work out energy levels of He⁺ using Bohr model. This also needed to review Bohr model for Hydrogen like atoms.

C: Problems on Newton’s Laws

5.  Two masses M₁ = 1 kg and M₂ = 3 kg are connected with a light string. The system is placed on an inclined plane of inclination 30⁰, such that the connecting string is parallel to the plane. The system slides down with M₂ pulling M₁. The coefficient of friction between M₁ and the plane is 0.2 while between M₂ and the plane is 0.1. Calculate the tension in the connecting string and the acceleration of the system? [JEE 1979]

6.   In above problem (5) what if M₁ and M₂ are interchanged?

Tasks involved in these problems (5 & 6) are:

1.      To identify the body in motion and forces acting on it.

2.      To resolving these forces in to rectangular components, one parallel and other perpendicular to motion.

3.      To applying Newton’s 3^rd law for reaction forces.

4.      To find net force in the direction of motion and applying Newton’s 2^nd law of motion.

5.      To draw free body diagram. (Halliday 2005)

The following is the smaller problem used.

A 5.1   A block kept on a simple inclined plane: (i) without and (ii) with friction. This is a standard problem to find acceleration of the block depending on angle of inclination.

These simple problems teach students the method of applying Newton’s laws of motion to simple system to study motion.

7.^*   A small ball is suspended from point A by a thread of length l. A nail is driven into the wall at a distance l/2 vertically below A at O. The ball is drawn aside so that the thread takes up a horizontal position and then released. At what point in the ball’s trajectory, will the tension in the thread disappear? What will be the highest point to which it will rise?

• This problem deals with motion in a vertical circle. This is also an application of Newton’s law of motion, for motion which is along circular path. The problem also includes gravity. This problem illustrates how centripetal force is responsible for motion specially in this case when it is not constant and the agency responsible for it.

Tasks involved in this problem are:

1. To draw a free body diagram for a particle moving along circular path.
2. To apply condition for particle to continue along circular track. In this case it is the conservation of energy.
3. To work out mathematical condition in part (b).
4. To understand how motion in a vertical circular track is different from uniform circular motion.
5. To find subsequent motion, after the condition stops being satisfied?

The following are the smaller problems used.

A 7.1 A bob of mass m is suspended by a light string of length L. It is imparted horizontal velocity v_o at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the top most point C. (i) Obtain an expression for v_o ; (ii) the speed at points B when string is horizontal and point C; (iii) the ratio of the kinetic energies  at B and C. Comment on the nature of the trajectory of the bob after it reaches the point C. [NCERT 2006]

8.   In Atwood’s machine, one block has a mass of 500 g and the other a mass of 460 g. The pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.0 cm. When released from rest the heavier block is observed to fall 75 cm in 5.0 s. What is the rotational inertia of the pulley? (Halliday 2005)

• Traditionally inclined plane and a pulley have always been treated as touchstone problems (Redish 1994) as this is where students learn resolution of forces, appropriate direction, use of Newton’s 2^nd law, use of force of friction, how to handle tension and reaction forces. This problem lays foundation of application of Newton’s Laws of motion.

Tasks involved in this problem are:

6.      To identify the body in motion and forces acting on it.

7.      To resolving these forces in to rectangular components, one parallel and other perpendicular to motion.

8.      To applying Newton’s 3^rd law for reaction forces.

9.      To find net force and torque in the direction of motion and applying Newton’s 2^nd law of motion.

10.  To draw free body diagram.

The following is the smaller problem used.

A 8.1  An ideal pulley with two masses hanging at the end of an ideal string. How would the system move if the two masses are unequal? This too is a standard problem.

9.  A ladder is resting equally inclined between a rough vertical wall and ground. If a man whose weight is half that of the ladder starts ascending it, show that the man can climb up to 71.4 % of the length of the ladder before it just starts slipping. Also show that this man can just make it to the top of the ladder if a boy of weight 1/4^th that of the ladder stands against the ground end of the ladder. Take the coefficient of friction between the ladder and (i) the ground and (ii) wall as 1/2 and 1/3 respectively*.

    This is a problem of static equilibrium and students need to identify and balance all the forces and torques.

Tasks involved in this problem are:

1. Identify the forces and torques acting on various parts of the bodies involved.
2. Resolve forces into horizontal and vertical components.
3. Identify clockwise and anti-clockwise moments.
4. In equilibrium net force and net moment should be zero to get the desired information.

The following is the smaller problem that may be used.

A 9.1 As shown in figure below, the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE, 0.5 m is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by floor on the ladder. [Hint: Consider the equilibrium of each side of the ladder separately.] [NCERT 2006]

10. A rod of mass M and length L is suspended horizontally by means of two massless strings of equal lengths attached at its end points. Find the tension in one of the strings at the moment the other is just cut*.

• This is also a touch stone problem in the same sense as inclined plane as it is a first exposure to deal with rotational motion and application of Newton’s laws to rotational motion.

Tasks involved in this problem are:

1. To draw diagram representing all the forces.
2. To resolve forces in to components to apply conditions for translational and rotational motion.
3. To find mathematical condition using Newton’s laws and solve the equation to get desired result.

Solutions:

4. Collision:

(i) 

       Conservation of momentum :

       In x-direction   :                        m_nv_o = m_Hev_He cosq  Þ    v_o = 4v_He cosq   

       In y-direction   :                        m_nv_n = m_Hev_He sinq  Þ    v_n = 4v_He sinq   

       Conservation of energy:       ½ m_nv_o² = ½ m_nv_n² + ½ m_Hev_He² + (E_n^(He+) – E₁^(He+))

                                                               Where last term in the bracket indicate excitation of helium due to inelastic nature of the process. This also indicates that K.E. is not conserved in the process.

        Here    E_n^(He+) =  4 E_n^(H)  =  -54.4/n²  eV

      Solving these equations we get :  ½ m_nv_n² =  - 4.52  +43.52/n²  eV

      Thus allowed K.E. of neutrons are :

        ½ m_nv_n² = 39 eV (for elstic collision-no excitation),  6.36 eV  (1^st excitation of He⁺)          

                         0.3156 (2^nd excitation He⁺)

(ii) Energy levels of He ⁺ are  G.S.  E₁ = -54.4 eV,  (1^st excited state)E₂ = -13.6 eV  and

     (2^nd  excited state)E₂ = -6.04 eV  

     Corresponding frequencies that can be emitted are  9.846´10¹⁵ Hz, 1.167´10¹⁶ Hz  and 1.823´10¹⁵ Hz  (n_21, n₃₁  and n₃₂ respectively).

5. Newton’s law:

m₂g sinq - mm₂g cosq - T = m₂a

m₁g sinq +T - mm₁g cosq = m₁a

  Solving which we get, 

                          and       

6. Newton’s law:

In problem 5, if masses are interchanged, they come in contact and T gets replaced by R, contact force between the blocks.

7. Newton’s law:

Total energy at A  :       mgl   ( where l is length of the string).   Height measured w.r.t. B.

At D where tension becomes zero :  total energy :   ½ mv² + ½ mgl(1 + cosq )

At D condition for tension becoming zero :      : centripetal force

Applying energy conservation and solving above equations we get         

At D particle becomes like a projectile projected at an angle q to the horizontal with speed  that we get by substituting cosq in centripetal force equation.

The maximum height reached by the ball above D will therefore be

Thus the heighest  point reached by the ball   is at  h =  l/2 + l cos q  + H = 5l/6 + 5l/54

                                                                                    =    above B   (or  below A)

8. Newton’s law:

m₁g – T₁ = m₁a ,       T_2­ – m_2­g = m₂a          and       R(T₁ – T₂)= Ia

with            a = Ra  = 2S/t²                 gives the needed answer.

9.Static Equillibrium:

Given: m_g = 1/2  and m_w = 1/3 .

Since ladder is equally inclined, q = 45^o,  let length of the ladder be l  and let the man climb up to l’ before ladder starts slipping.

In equilibrium, net force in any direction and net moment of force is zero.

Since no net horizontal force:     N₁ + m_w N₂ =  3W/2 

Since no net vertical force:                      N₂  = m_g N₁

Solving which we get    N₁ = 9W/7    and    N₂ = 9W/14

(i)  Just before the ladder slips: no net moment: 

[W(l/2) + (W/2)l’] cos q  = N₂ (1 + m_w)l sin q

                       Solving which we get               l’ = (5/7)l  = (0.7143) l

(ii) If the boy of weight W/4 is added at the ground end the equation for vertical force changes to N₁ + m_w N₂ =  7W/4   and  this gives    N₁ = 3W/2    and    N₂ = 3W/4

With new value of N₂   we get        l’ = l

10. Rotational Motion :

The moment at which scissor cuts one of the string,

net downward force produces acceleration of centre of mass :  

                                          Mg – T = M a

Net torque produces rotational acceleration around the other end of the rod  

                                                           a = a/(l/2)     

                                         \       Mg(l/2) =  Ia

where I = Ml ²/3 is MI of rod around one end.  

              Solving this we get              T = Mg/4     and  a = 3g/4

References:

       *    Source unknown

1. Halliday, Resnick and Walker, Fundamentals of Physics by 6^th Ed., John Wiley & Sons (2005)
2. JEE 1993- Joint Entrance Examination for admission to IIT -1993

3. Mody A. K. & Pradhan H. C., ‘Problem Based Learning in Basic Physics – I, School Science 49 (3) Sept 2011

4. Pradhan H.C. & Mody A. K., ‘Constructivism applied to physics teaching for                                     capacity building of undergraduate students’, University News, 47 (21) 4-10, (2009)

5. Physics Textbook for class XI part-I, NCERT New Delhi (2006)