Today I give you a free response practice question on kinematics. I’ll give you the answer in my next post.
A ball at a height H is released from rest at the instant t = 0 and it strikes a fixed touch plate P, inclined at 45º with respect to the vertical, at the instant t = t1. The point where the ball strikes the plate is at a height h (fig.). The collision between the ball and the plate is elastic. After the collision the ball continues its motion and strikes the ground at the instant t = t2. Now, answer the following questions:
(a) If the height h where the ball strikes the plate is varied, how will you indicate, graphically, the variation of t1 with (H – h)? Use the coordinate axes as shown in the figure below to show the nature of variation.
(b) After the collision what is the nature of the path followed by the ball? Pick out your answer from the following three options by putting a tick (√) mark against the correct one:
Straight line path ……
Parabolic path ……
Arc of a circle …..
Justify your answer.
(c) Derive an expression for the total time of flightt2 of the ball.
(d) Determine the horizontal displacement of the ball when it strikes the ground.
AP Physics C aspirants should answer the following part also:
(e) For a given value of H determine the value of h for which the time of flight t2 of the ball is maximum.
Try to answer the above question. I’ll be back soon with a model answer for your benefit.
Example isn't another way to teach, it is the only way to teach.
– Albert Einstein
Questions (Multiple Choice and Free Response) involving rotational motion were discussed on various occasions on this site. Essential formulae to be remembered in this section were discussed in the post dated 20th January 2008. You can access all the posts related to this section by clicking on the label ‘rotational motion’ below this post or by trying a search using the search box provided on this page. In both cases you need to use ‘older posts’ links to obtain all related posts.
Some of the questions on rotational motion may prove to be some what difficult and time consuming for many among you. Once you master the basic principles, you will become more confident. Today we will discuss a few more multiple choice practice questions involving rotational motion:
(1) A metre stick of mass 0.2 kg can be balanced on a knife edge at the 60 cm mark (see figure) when a piece of rock (specific gravity = 3) fully immersed in water is suspended from the 70 cm mark. If the piece of rock were in air, suspended from the 70 cm mark itself, what would be the position of the knife edge for the balance?
(a) 56 cm
(b) 58 cm
(c) 62 cm
(d) 64 cm
(e) 66 cm
The weight of the metre stick acts through its centre of gravity, which is at the 50 cm mark. The torque due to the weight of the metre stick tries to rotate it in the anticlockwise sense where as the torque due to the apparent weight of the piece of rock in water tries to rotate the metre stick in the clockwise sense. These torques have equal magnitudes since the metre stick is horizontal. Since the lever arms are equal (0.1 m each), the apparent weight of the piece of rock must be equal to the weight of the metre stick which is equal to 0.2 kgwt.
[Note that we got the above result by equating the magnitudes of the torques: 0.2×0.1 = w2×0.1 where w2 is the apparent weight (or, the weight inside water) of the piece of rock].
If the weightof the piece of rock in air is w1, the loss of weight in water is w1 – w2 and the specific gravity, which is equal to 3 is given by
w1/( w1 – w2) = 3
Or, w1/( w1 – 0.2) = 3
Therefore, w1 = 0.3 kgwt.
When the piece of rock is in air, the knife edge should be at distance x from the centre of gravity such that
0.2×x = 0.3×(0.2 – x)
This gives x = 0.06/0.5 = 0.12 m = 12 cm
The position of the knife edge is therefore 50 cm + 12 cm = 62 cm.
(2) A sphere rolls without slipping along a horizontal surface. If the velocity of the centre of mass of the sphere is v, what is the velocity of the point of contact (with the horizontal surface) of the sphere?
(a) 2 v
(b) v
(c) v/2
(d) v/4
(e) zero
When the sphere rolls, it has an angular motion (rotation) and a linear motion (translation). If the the horizontal surface is perfectly smooth, the sphere will slip fully and will just spin about its central axis. In other words, it will have angular motion alone and the velocity of the contact point will be v, directed backwards.
When the surface is rough, the sphere rolls forward and the centre of mass moves forward with velocity v. Since the sphere is rigid, the contact point (lowest point) of the sphere too has to move forward with the velocity v. Since the contact point has an additional backward velocity v due to the spin of the sphere, the resultant velocity of the contact point on the sphere is zero [Option (e)].
[In the above question what is the velocity of the top point of the sphere? You can easily show that it is 2 v, forward].
(3) For opening a door 1 m wide, the minimum force required is 20 N.If the door is pushed at a distance of 0.4 m from the line of the hinges, what is the minimum force required for opening it?
(a) 8 N
(b) 10 N
(c) 20 N
(d) 40 N
(e) 50 N
The force required for opening the door will be minimum if it is applied perpendicular to the plane of the door, with the point of application of the force farthest away from the line of the hinges (at the edge of the door, opposite to the line of the hinges). This ensures that the torque produced by the force is maximum. Evidently the minimum torque required for opening the door is 20×1 Nm = 20 Nm.
[Remember that torque = force×lever arm. It is the torque (and not the force) that matters for opening the door].
If the force is applied at a distance of 0.4 m from the line of the hinges, the minimum force F required for opening the door is given by
F×0.4 = 20 Nm
Therefore, F = 50 N.
The following questions are specifically for the AP Physics C aspirants:
(4) The angular displacement θof a fly wheel at the instant t is given by
θ =2t3– 4 t2 + 8
The angular acceleration of the fly wheel after 2s in is
(a) 20 rad/s
(b) 16 rad/s
(c) 12 rad/s
(d) 8 rad/s
(e) 6 rad/s
The angular acceleration α is given by
α = d2θ/dt2
Since dθ/dt = 6 t2– 8 t, we have
α = 12 t – 8
The angular acceleration after 2 s = (12×2)– 8 = 16 rad/s
(5) A cylinder of moment of inertia 2 kgm2 has angular displacement θ given by
θ =3t2– 4 t
The torque acting on the cylinder
(a) decreases linearly with time
(b) increases linearly with time
(c) decreases non-linearly with time
(d) increases non-linearly with time
(e) remains constant
The angular acceleration of the cylinder is α = d2θ/dt2 = 6 rad/s2.
The torque τ is given by
τ = I α where I is the moment of inertia.
Therefore, τ =2×6 = 12 Nm
Since this is constant (independent of time), option (e) is correct.
[The expression for the angular displacement θ is similar to the expression s = ut + ½ at2 for the linear displacement in uniformly accelerated motion. Therefore from the form of the expression itself you should be able to understand that the angular acceleration and hence the torque, is constant].
(6) Two identical solid hemispheres, each of mass mand radius r are welded together as shown. What is the moment of inertia of this system about the axis AB which is perpendicular to the plane surfaces of the hemispheres? [Moment of inertia of a solid sphere of mass Mand radius Rabout any diameter is (2/5)MR2]
(a) (14/5)mr2
(b) (7/5)mr2
(c) (6/5)mr2
(d) (14/5)mr2
(e) (14/5)mr2
Since the moment of inertia of a solid sphere of mass Mand radius Rabout any diameter is (2/5)MR2, its moment of inertia I about a tangent is (by the theorem of parallel axes) given by
I = (2/5)MR2 + MR2 = (7/5)MR2
The moment of inertiaof a hemisphere about a tangent such as AB must be half of this. Since the axis AB is the common tangent to the two hemispheres, the total moment of inertia of the system containing the two hemispheres must be (7/5)MR2.
In the above expression M is the mass of the entire sphere and we have
M = 2m
Also R = r as given in the question.
Therefore, the answer is (7/5)(2m)r2 = (14/5)mr2
You will find few more practice questions (with solution) in this section here.
Questions involving magnetic fields posted on this site earlier can be accessed by clicking on the label ‘magnetic field’ below this post or by trying a search for ‘magnetic field’ using the search box provided on this page. Answering questions on magnetic field will not be generally difficult; but occasionally you will find questions consuming too much of your precious time. Today we will discuss a few more multiple choice questions involving magnetic fields:
(1) Two straight long vertical wires W1 and W2 carry steady currents I and 3I respectively as shown. P is a point midway between the wires. The resultant magnetic flux density at P due to the currents in the wires is B. The direction of B is
(a) vertically upwards
(b) vertically downwards
(c) horizontal and directed normally into the plane of the figure (away from the reader)
(d) horizontal and directed normal to the plane of the figure, towards the reader
(e) horizontal and directed towards the wire W2
The magnetic field at P due to the current I in the wire W1 is directed normally into the plane of the figure (away from the reader). The magnetic field at P due to the current 3I in the wire W2 opposes the above field. Since the magnitude of the field due to the wire W2 is three times that due to the wire W1, the resultant field will be directed normal to the plane of the figure, towards the reader [Option (d)].
(2) In the above question if the current 3I in the wire W2 is switched off, what will be the magnetic field at P?
(a) B
(b) – B
(c) – B/2
(d) B/2
(e) – B/3
The direction of the magnetic field produced by the current I in the wire W1 is opposite to the net field B in question No.(1). So the sign has to be negative. Since the currents I and 3I in the wires W1 and W2 produce opposing fields at P (in question no.1), the resultant field B has magnitude proportional to 2I. When the current 3I in the wire W2 is switched off, the field at P is proportional to I so that the magnitude is reduced to half the net value in question No.(1).
The field at P in this case is therefore – B/2.
The following questions are specifically meant for AP Physics C aspirants:
(3) The magnetic flux density at the centre of a plane circular coil of radius R carrying a current I is B0. If the current in the coil is reversed, the magnetic flux density on the axis of the coil atdistance R from the centre will be
(a) B0/√8
(b) – B0/√2
(c) – B0/√8
(d) – B0/2
(e)B0/2
The magnitude of the magnetic flux density B at a point on the axis of the coil atdistance x from the centre of the coil is given by
B = μ0nR2I/2(R2 + x2)3/2 where n is the number of turns in the coil.
The magnetic field B0 at the centre (x = 0) of the coil is given by
B0 = μ0nI/2R
The field (B1) on the axis at distance R from the centre (x = R) is given by
B1 = μ0nR2I/2(R2 + R2)3/2 = μ0nR2I/4√2 R3
Or, B1 = μ0nI/4√2 R
Therefore, B1 = B0/2√2 = B0/√8
Since the current in the coil is reversed, the direction of the magnetic field must be reversed so that the correct option is – B0/√8.
(4) An electron moving with velocity (30 i +40 j) ms–1 enters a magnetic field of – 2 k tesla where i, j and k are unit vectors along the x, y and z directions respectively. Then
(a) the speed of the electron will not change; but the path will become parabolic
(b) the speed alone will change
(c) the speed will not change; but the path will become helical
(d) the speed will change and the path will become circular
(e) the speed will not change; but the path will become circular
Outside the magnetic field the path of the electron is straight and is contained in the xy plane. The magnetic field is directed along the negative z-direction an hence the electron enters normally into the magnetic field. A magnetic force acts perpendicular to the direction of motion of the electron.
The path of the electron within the magnetic field will therefore be circular. But the speed will be unchanged since a stationary magnetic field cannot change the kinetic energy of the electron. The correct option is (e).
[If the initial velocity of the electron had an z-component also (for instance, v = 30 i +40 j + 20 k), the electron would enter the magnetic field at an angle other than 90º and then the path would be helical].
