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`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Sunday, September 14, 2008

### Answers to Free Response Practice Questions on AP Physics Kinematics

In the post dated 10th September 2008, two free response practice questions were given to to without the answer. As promised, I give below the answers along with the questions:

(1) Points A, B and C lie on a straight line parallel to the X-axis in a region of space where a small uniform electric field E directed along the negative X-direction exists. Other fields (including gravitational field) are negligible. A proton of mass m and charge e is projected from point B with velocity u along the positive X-direction. A………….B………………..C

Now, answer the following questions:

(a) Draw a graph to indicate qualitatively the nature of variation of the displacement of the proton with time from the instant of projection to the instant it returns to the point of projection. (Take the time t along the X-axis and the displacement s along the Y-axis). Explain why the shape of the graph is as shown by you.

(b) Draw a graph to indicate qualitatively the nature of variation of the velocity of the proton with time from the instant of projection to the instant it returns to the point of projection. (Take the time t along the X-axis and the velocity v along the Y-axis). Explain why the shape of the graph is as shown by you.

(c) Determine the time T required for the proton to attain the maximum displacement BC and indicate this time T in the velocity–time graph

(d) Determine the maximum displacement BC

(e) Another proton was projected simultaneously from point B with the same speed u along the negative X-direction. The first proton arrived at the point A in time t1 and the second proton in a shorter time t2. If a third proton is released (from rest) at the point B, determine the time required for it to reach the point A.

(a) The required displacement–time graph is shown in the figure. This is a case of uniformly accelerated one dimensional motion. The proton being positively charged, the acceleration of the proton is directed along the negative X-direction (which is the direction of the electric field). Acceleration has magnitude Ee/m which is constant. The magnitude of the velocity of the proton goes on decreasing. At the point C the velocity becomes zero and then gets reversed. Thereafter the magnitude of the velocity (along the negative X-direction) goes on increasing. The displacement-time graph is therefore non linear as indicated.

(b) The velocity-time graph is linear since the acceleration is constant. At the instant of projection the velocity is positive and has magnitude u. At C the velocity is zero as explained above. The lower portion of the graph shows the reversal of the velocity and the increase in its magnitude linearly in the opposite direction.

(c) Since v = v0 + at where v, v0 and a are respectively the final velocity, initial velocity and acceleration, we have

0 = u – (Ee/m)T from which T = um/Ee

The time T is indicated in the velocity-time graph

(d) When the displacement is maximum, the velocity v of the proton is zero. Therefore, we have v2 = u2 2ax where x is the maximum displacement. Thus 0 = u2 2(Ee/m)x

From this x = u2m/2Ee

(e) The displacements of the two protons when they arrive at A are the same. Therefore we have

ut1 – (½)at12 = – ut2 – (½)at22. Here a is the common acceleration (Ee/m) which is in the negative X-direction. The initial velocity of the second proton also is in the negative X-direction.

The proton released at B from rest also has the same displacement when it arrives at A. Therefore, if t is the time taken by it to reach the point A, we have

ut1 – (½)at12 = – (½)at2 and

ut2 – (½)at22 = – (½)at2

Dividing, t1/t2 = (t12 t2)/( t2 t22)

This yields t = √(t1t2)

(2) An iron ball of mass m released from rest at time t = 0 from a stationary balloon at a height falls under gravity which can be assumed to be constant. While falling down, the ball experiences a viscous drag force D (due to the air) in the form D = bv where v is the velocity of the ball and b is a constant. Now answer the following questions in this context:

(a) Assuming that the acceleration due to gravity g is constant throughout the path of the ball, draw a graph to indicate the nature of variation of the acceleration of the ball with its velocity. Take the velocity v along the X-axis and the acceleration a along the Y-axis. Incorporate all possible values of velocity in the graph and give the reason for the shape of the graph.

(b) Write a differential equation for the acceleration of the ball.

(c) Solve the differential equation you have written in part (b) to obtain the time-dependent velocity of the ball in terms of the given parameters and fundamental constants.

(d) From the expression for the velocity obtained in part (d) obtain the terminal velocity of the ball.

(e) If the ball were moving through water instead of air, how will the terminal velocity be affected? Put a tick mark against the correct statement out of (i), (ii) and (iii) given below:

(i) Terminal velocity will be unchanged ___

(ii) Terminal velocity will be increased ___

(iii) Terminal velocity will be decreased ___

Justify your answer giving two important reasons.

(a) The net force on the ball is mg bv, taking the downward gravitational force mg as positive. The drag force bv is opposite to the velocity v and is therefore negative.

Acceleration of the ball is a = (mg bv)/m = g – (b/m)v

Initially the velocity is zero and there is no viscous drag so that the acceleration is equal to g. As v increases, the acceleration a decreases linearly and when g = (b/m)v, the acceleration becomes zero. (This is the case of the magnitudes of gravitational pull and the viscous drag becoming equal). The ball then moves with a constant velocity (Terminal velocity). The velocity cannot increase beyond this value since the net force is zero.

The required graph is shown in the adjoining figure

(b) The differential equation for acceleration is dv/dt = g – (b/m)v

(c) To obtain v, integrate the above equation:

∫[dv/(g – (b/m)v)] = dt

This gives – (m/b)ln[g – (b/m)v] = t + C where C is the constant of integration which can be found from the initial conditions.

We have v = 0 when t = 0. Substituting these in the above equation, C = – (m/b)ln g

Substituting for C in the above equation,

– (m/b)ln[g – (b/m)v] = t – (m/b)ln g

Rearranging, ln[1 – (b/mg)v] = – (b/m)t

Therefore, 1 – (b/mg)v = e– (b/m)t so that

v = (mg/b)[1– e– (b/m)t]

(d) The terminal velocity (vT) is attained when the time t is sufficiently large so that the value of e– (b/m)t in the above expression for v becomes negligible. Therefore,

vT = mg/b

[You can easily obtain the value of vT by equating the magnitudes of the gravitational pull and the viscous drag: mg = b vT so that vT = mg/b]

(e) If the ball were moving through water instead of air, the terminal velocity would be decreased. This happens because of two important reasons:

(1) The viscosity of water is much greater than that of air. In other words, the constant b is much greater.

(2) The force of buoyancy in water is very much significant where as that in air is negligible. The force of buoyancy counteracts the gravitational pull, thereby reducing the acceleration of the ball. In other words, in the expression vT = mg/b, the real weight mg of the ball is to be substituted by the reduced apparent weight.