Newton’s Laws of Motion for AP Physics- Equations to be Remembered

We shall require a substantially new manner of thinking if mankind is to survive.

– Albert Einstein

Even though most of you will be remembering the important points in connection with Newton’s Laws of Motion, it will be better to have a glance at the following:

(1) Inertia is a basic property of any material body, by virtue of which it resists any change in its state of rest or of uniform motion.

(2) A force is required to change the state of rest or of uniform motion of a body. The resultant force acting on a body at rest or in uniform motion is zero. [Note that a body in uniform motion has uniform velocity].

(3) Newton’s second law is mathematically expressed as

F_net = ma where F_net is the net (resultant) force and a is the acceleration.

This can be written in terms of momentum p as

F_net = dp/dt

Often we write this as F= dp/dt, understanding that F is indeed the net force.

[Remember that the mass m can be treated as constant only at speeds negligible compared to the speed of light].

(4) Impulse given to an object (by a force) = F∆t where F is the force and ∆t is the time for which the force acts.

If the force is not constant and it acts from the instant t₁ to the instant t₂, we have

Impulse = _t1 ∫^t2 Fdt

This gives the area under the force-time graph between the ordinates corresponding to the times t₁ and t₂ (Shaded area in fig.)

Since force F = ∆p/∆t where ∆p is the change in momentum during the time ∆t, we can write

Impulse = F∆t = (∆p/∆t) ∆t = ∆p

Thus impulse = change of momentum

(5) Motion in a lift

The weight of a body of mass ‘m’ in a lift can be remembered as m(g-a) in all situations if you apply the proper sign to the acceleration ‘a’ of the lift. The acceleration due to gravity ‘g’ always acts vertically downwards and its sign may be taken as positive. The following cases can arise in this context:

(i) Lift moving down with acceleration of magnitude ‘a’:

In this case ‘a’ also is positive and the weight is m(g-a) which is less than the real weight of the body (when it is at rest).

(ii) Lift moving up with acceleration:

In this case ‘a’ is negative and the weight is m[g-(-a)] = m(g+a).

(iii) Lift moving down with retardation (going to stop while moving down):

In this case also ‘a’ is negative and the weight is m[g-(-a)] =m(g+a) which is greater than the actual weight.

(iv) Lift moving up with retardation (going to stop while moving up):

In this case ‘a’ is positive and the weight is m(g-a)

(v) Lift moving up or down with uniform velocity:

In this case ‘a’ is zero and the weight is mg.

(vi) Lift moving down with acceleration of magnitude ‘g’ (falling freely under gravity as is the case when the rope carrying the lift breaks):

In this case ‘a’ is positive and the weight is m(g-g) which is zero.

If you have a clear idea of the weight of a body in a lift, you will be able to use it in other similar situations as well (for instance, the motion of bodies connected by a string passing over a pulley).

[We will discuss conservation of momentum separately in due course].

(6) Friction

The force of friction, F_fric ≤ μN where μ is the coefficient of friction and N is the normal reaction (normal force).

In the adjoining figure, a body being pulled along a horizontal surface by a horizontal force F is shown. The frictional force F_fric is maximum when the body just begins to move and is called limiting force of static friction (F_s)_max so that we have

(F_s)_max = μ_s N. This gives the value of the coefficient of static friction μ_s as

μ_s = (F_s)_max /N

When the body slides along the surface, the friction called into play is called kinetic friction. The force of kinetic friction F_k is less than the above limiting value (F_s)_max and the corresponding coefficient of kinetic friction μ_k is less than μ_s. We have

μ_k = F_k/N

If the body rolls along the surface, The friction called into play is called rolling friction which is much less than kinetic friction.

Angle of friction λ is the angle between the the normal force N and the resultant reaction S. As shown in the figure, the resultant reaction is the resultant of the normal force N and the frictional force F_fric. Since tan λ = F_fric/N, it follows that

μ = tan λ

A body of mass m placed on a ramp (inclined plane) is shown in the adjoining figure. The component mg sinθ of the weight mg of the body is the force trying to move the body down the plane. The normal reaction is the reaction (force) opposing the component mg cosθ of the component of the weight of the body normal to the inclined plane. The frictional force F_fric is opposite to the component mg sinθ (of the weight of the body) parallel to the plane. Note that friction is a self adjusting force up to its maximum value (F_s)_max and if the body shown in the figure is at rest, F_fric is just sufficient to balance the component mg sinθ (of the weight of the body).

If the inclination of the plane is gradually increased from a small value, the body placed on it will begin to slide down when the angle is equal to the angle of friction, λ. The angle of repose is therefore equal to the angle of friction.

Often you may be asked to draw a free body diagram (FBD), indicating the forces acting on the body. In the case of the body placed on the inclined plane, the free body diagram is as shown. The body is represented by a dot. The forces to be shown are the weight mg of the body, the normal force N (equal to mg cosθ) exerted by the inclined surface on the body and the frictional force F_fric since they are the actual forces acting on the body. Don’t worry about the components mg sinθ and mg cosθ of the weight. The real force is the weight mg which we have shown already. We consider the components just for the convenience of explanation. The normal reaction (force) offered by the surface and the frictional force between the body and the surface are to be accommodated in addition to the weight of the body.

[Note that if you want, you can draw the FBD showing the components mg sinθ and mg cosθ of the weight of the body. But in that case you will not show the weight mg in the FBD].

If the inclined plane is smooth, the frictional force F_fric will be absent in the free body diagram.

If the body is held on a smooth incline by a spring fixed to the incline, the spring force Kx has to be shown in place of the frictional force F_fric. Here K is the spring constant and x is the elongation (or compression as the case may be) of the spring.

If the body moves down the incline and the viscous drag force (air resistance) is significant, that too is to be shown up the incline.

In the next post we will discuss questions in this section. Meanwhile, find some useful multiple choice questions (with solution) here.

Answers to Free Response Practice Questions on AP Physics Kinematics

In the post dated 10^th 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 (v_T) 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,

v_T = mg/b

[You can easily obtain the value of v_T by equating the magnitudes of the gravitational pull and the viscous drag: mg = b v_T so that v_T = 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 v_T = mg/b, the real weight mg of the ball is to be substituted by the reduced apparent weight.

Free Response Practice Questions on AP Physics Kinematics

As in the case of multiple choice questions, free response questions in the AP Physics Exams are designed to test your knowledge, understanding and capability for applying the things you have studied. But unlike the multiple choice questions, the presentation of the answers of free response questions is very important since you have to give all the required details in an effective manner within the stipulated time.

Here are two free response practice questions on kinematics. Question No.1 is for AP Physics B; but it will be useful for AP Physics C aspirants also. Question No.2 is for AP Physics C only.

(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.

(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 would 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.

The above questions carry 15 points each. You can take about 17 minutes for answering question No.1 and about 15 minutes for answering question No.2. Try to answer these questions. I’ll be back soon with model answers for your benefit.

AP Physics B and C – Additional Multiple Choice Questions on Two Dimensional Motion

Everything that is really great and inspiring is created by the individual who can labor in freedom.
– Albert Einstein

As promised in the post dated 27^th August 2008, I give below some more multiple choice questions (with solution) on two dimensional motion. These questions will be useful for AP Physics B as well as AP Physics C aspirants.

(1) A particle is projected from the origin with velocity (2 î + 3 ĵ) ms^–1 where î and ĵ are unit vectors along the X and Y directions respectively. If the X direction is horizontal, what is the range of the projectile on the horizontal plane through the point of projection? (g = 10 ms^–2)

(a) 6.2 m

(b) 4.8 m

(c) 4.2 m

(d) 2.8 m

(e) 1.2 m

The time of flight of the projectile can be found from the vertical component of velocity (which is 3 ms^–1). This projectile will have time of flight equal to that of a particle projected vertically up with velocity 3 ms^–1. Therefore, time of flight,

T_f = 2 v_y/g = 2×3/10 = 0.6 s

[We have used the relation s =ut + ½ at² in which s = 0, u = 3 ms^–1, a = g and t = T_f]

Horizontal range = Horizontal velocity ×Time of flight = 2×0.6 = 1.2 m

(2) The horizontal rnge R and the maximum height H (both in metre) of a projectile are related to the time of flight T (in seconds) as

R = 40T and H = 15T – 1.25T²

The velocity of projection is (g = 10 ms^–2)
(a) 100 ms^–1

(b) 50 ms^–1

(c) 40 ms^–1

(d) 30 ms^–1

(e) 25 ms^–1

Projectile motion is the combination of two one dimensional motions. From the form of the expressions for R and H it is evident that the horizontal component of the velocity of projection is 40 ms^–1. [Remember that R = v_0xT where v_0x is the horizontal component of the velocity of projection]

The vertical component of the velocity of projection is 30 ms^–1 [since the maximum height, H = v_0y(T/2) – ½ g(T/2)²]

The velocity of projection is √(v_0x² + v_0y²) = √(40² + 30²) = 50 ms^–1.

(3) A body of mass 2 kg is projected from the origin with initial velocity v = (40 î + 30 ĵ) ms^–1. The only force acting on the body is a constant force F = 3 î – 4 ĵ newton. The time in which the y–component of the velocity of the body will become zero is

(a) 5 s

(b) 10 s

(c) 15 s

(d) 20 s

(e) 25 s

The y–component of acceleration is – 4/2 = – 2 ms^–1 [since the y–component of force is

– 4 N and the mass is 2 kg].

The time t in which the y–component of the velocity (30 ms^–1) will become zero is given by

0 = 30 –2t (using v = u + at)

This gives t = 15 s.

The following questions are specifically meant for AP Physics C aspirants:

(1) The position co-ordinates of a particle projected from the origin are given by x = 30t and y = 40t – 5t². The vertical component of velocity after 5 seconds and the magnitude of the velocity of projection are respectively

(a) –10 ms^–1, 50 ms^–1

(b) –10 ms^–1, 40 ms^–1

(c) 10 ms^–1, 40 ms^–1

(d) –20 ms^–1, 50 ms^–1

(e) 20 ms^–1, 50 ms^–1

The vertical component (v_y) of velocity at the instant t during the flight of the projectile is given by

v_y = dy/dt = 40 – 10t

When t = 5 s we have v_y = 40 – 10×5 = –10 ms^–1

[The negative sign shows that the projectile is moving down]

The initial vertical component of velocity (v_0y) at the instant when t = 0 is 40 ms^–1. The horizontal component of velocity at the instant of projection (and throughout the motion) is 30 ms^–1 since x = 30t and dx/dt = v_x = v_0x = 30 ms^–1. [Since x = 30t, you should be able to understand that the horizontal velocity is 30 ms^–1, even without finding the value of dx/dt]

The velocity of projection has magnitude √(v_0x² + v_0y²) = √(30² + 40²) = 50 ms^–1.

(2) The displacement of a particle at time t has x-component 3 + at³ and y-component 4 + bt³. The speed of the particle at time t is

(a) 3t²√(a ² + b²)

(b) 3t²(a ² + b²)

(c) 3t√(a ² + b²)

(d) 3t(a ² + b²)

(e) t²√(a ² + b²)

The x-component of velocity (v_x) is the time derivative of the x-component of displacement: v_x = 3 at²

Similarly, y-component of velocity, v_y = 3 bt²

The magnitude (v) of the velocity at time t = √(v_x² + v_y²)

Therefore, v =√[(3 at² )² + (3 bt²)²] = 3t²√(a ² + b²)

This is the speed of the particle at time t.