Here are some multiple choice questions designed to test your knowledge, understanding and capacity for application of Newton’s laws:
(1) A block of mass M equal to 50 kg is suspended from the ceiling by means of rope AB (fig.). Another block of mass m equal to 1 kg is suspended from the 50 kg block, by means of rope CD. The ropes AB and CD have the same breaking strength F. If the 10 kg block is pulled suddenly with a force slightly less than F, what will happen?
(a) Rope CD will break
(b) Rope AD will break
(c) Both AB and CD will break
(d) Neither AB nor CD will break
(e) Either AB or CD will break
This question is meant for checking your understanding of the property of inertia. The rope CD will certainly break since the pulling force F and the weight mg of the smaller mass m is applied on it. Because of the large inertia of the large mass M, movement of M cannot take place immediately. Since the string CD breaks, the required breaking force is not communicated to the rope AB so that it cannot break. The correct option is (a).
[If you pull the 1 kg mass slowly with a gradually increasing force, the rope AB will break. You should be able to explain this behaviour].
(2) A mother and her child (fig.) hold the ends of of a rope AB while standing at a distance of 10 m apart on a horizontal surface having negligible friction. The mother and the child have masses 70 kg and 30 kg respectively. They start pulling on the rope thereby trying to reduce the distance between them. What will be distances moved by the mother and the child if they reduce the distance between them by 5 m?
(a) Mother: 5m, Child: 0 m
(b) Mother: 2.5m, Child: 2.5 m
(c) Mother: 1.5m, Child: 3.5 m
(d) Mother: 3.5m, Child: 1.5 m
(e) Mother: 0 m, Child: 0 m (They cannot move)
The forces exerted by the mother and the child will be equal and opposite in accordance with Newton’s third law so that the rate of change of momentum (∆p/∆t)of the mother and the child will be equal and opposite. Since they reduce the distance in the same time ∆t, the change of momentum will be equal and opposite. If the distance moved my the mother is x, the distance moved by the child is 5 – x.
Equating the magnitudes of momenta of the mother and the child, we have
70×x/∆t= 30×(5 – x)/∆t, from which x = 1.5 m.
The distance moved by the child is 5 – 1.5 = 3.5 m.
(3) Bodies A, B, C and D of masses 4 kg, 3 kg, 2 kg and 1 kg are connected by light inextensible strings and pulled along a horizontal frictionless surface with a force of 20 N. The tension T2 in the string connecting the 3 kg and 2 kg masses is
(a) 7 N
(b) 14 N
(c) 6 N
(d) 8 N
(e) 20 N
The common acceleration of the system is = Force/Total mass moved = 20/(4+3+2+1) = 2
ms–2
Since the tension T2pulls the 4 kg and the 3 kg masses with the above acceleration, we have
T2 = Total mass moved × common acceleration = (4+3)×2 = 14 N.
(4) A lift of mass 200 kg starts moving down from rest. The variation of its velocity for 8 seconds is shown in the adjoining velocity–time graph. What is the tension in the supporting wire during the 7th second?
(a) 5500 N
(b) 5000 N
(c) 4500 N
(d) 6000 N
(e) 4000 N
You should note that the lift is moving down and decelerating during the interval from 6 s to 8 s. The weight of the lift in this condition is m[g-(-a)] =m(g+a) where m is the mass of the lift, a is the acceleration of the lift and g is the acceleration due to gravity. (See the post dated 24th September 2008).
From the graph, a = 2/(8 – 6) = 1 ms–2.
Taking g = 10 ms–2, the weight of the lift at the 7th second = 500×11 = 5500 N
This is the tension in the wire supporting the lift.
We will discuss more questions in this section in the next post.
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
Fnet = ma where Fnet is the net (resultant) force and a is the acceleration.
This can be written in terms of momentum p as
Fnet = dp/dt
Often we write this as F= dp/dt, understanding that F is indeed the netforce.
[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 ∫t2Fdt
This gives the area under the force-time graph between the ordinates corresponding to the timest1 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 theactual 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, Ffric ≤ 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 Ffric is maximum when the body just begins to move and is called limiting force of static friction (Fs)max so that we have
(Fs)max = μsN. This gives the value of the coefficient of static friction μsas
μs= (Fs)max/N
When the body slides along the surface, the friction called into play is called kinetic friction. The force of kinetic friction Fk is less than the above limiting value (Fs)maxand the corresponding coefficient of kinetic frictionμk is less than μs. We have
μk = Fk/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 resultantreaction S. As shown in the figure, the resultant reaction is the resultant of the normal force N and the frictional force Ffric. Since tanλ = Ffric/N, it follows that
μ = tanλ
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 mgcosθ of the component of the weight of the body normal to the inclined plane. The frictional force Ffric 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 (Fs)max and if the body shown in the figure is at rest, Ffric 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 mgcosθ) exerted by the inclined surface on the body and the frictional force Ffricsince 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 mgin the FBD].
If the inclined plane is smooth, the frictional force Ffric 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 Kxhas to be shown in place of the frictional force Ffric. 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.
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. Takethe 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 whenthe 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.
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.2is 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. Takethe 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.
(a) Rope CD will break
(2) A mother and her child (fig.) hold the ends of of a rope AB while standing at a distance of 10 m apart on a horizontal surface having negligible friction. The mother and the child have masses 70 kg and 30 kg respectively. They start pulling on the rope thereby trying to reduce the distance between them. What will be distances moved by the mother and the child if they reduce the distance between them by 5 m?
