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

## Sunday, February 14, 2010

### Answer to Free Response Practice Question on Kinematics in Two Dimensions for AP Physics C

A free response practice question on kinematics in two dimensions was given to you in the post dated 11th February 2010. As promised, I give below a model answer along with the question:

A chopper at an air base is rising vertically up with a velocity of 2 ms–1. When it is at an altitude of 50 m, a packet is thrown down with velocity 8 ms–1 (with respect to the chopper), making an angle of 60º with the vertical. Neglecting air resistance and assuming that g = 10 ms–2, answer the following questions

(a) Draw a diagram to show the nature of the path AB followed by the packet from the moment it leaves the chopper to the moment it hits the ground, as seen by an observer on the ground. Is the path straight, circular or parabolic?.

(c) Calculate the time taken by the packet to reach the ground.

(d) Determine the horizontal distance traveled by the packet before hitting the ground.

(e) Determine the velocity (magnitude as well as direction) with which the packet hits the ground.

(a) The path AB of the packet is shown in the following figure. The path is parabolic. (b) The packet will not move up initially. This is due to the fact that the vertical, downward component (8 cos 60º = 4 ms–1) of the velocity with which the packet is thrown down is greater than the upward velocity (2 ms–1) of the chopper.

[The situation is as though the packet is thrown down from a stationary chopper with vertical (downward) velocity component (4 ms–1 2 ms–1) = 2 ms–1 and horizontal velocity component 8 sin 60º = 4√3 ms–1].

(c) The time t taken by the packet to reach the ground can be found by considering its vertical motion. We have

s = ut + ½ at2 where the vertical displacement s = 50 m, vertical (downward) velocity component u = 8 cos 60º – 2 = 2 ms–1 and vertical (downward) acceleration a = g = 10 ms–2.

Therefore, 50 = 2t + ½ ×10t2.

Or, 5t2 + 2t – 50 = 0

This gives t = [–2 ± √(4 – 1000)] /10 = 2.968 s.

(d) The horizontal distance R traveled by the packet before hitting the ground is the product of the above time and the horizontal velocity component vhorizontal which is equal to 8 sin 60º (= 4√3 ms–1).

Therefore, R = 2.968×4√3 = 20.56 m.

(e) The velocity with which the packet hits the ground is the resultant of the horizontal and vertical components. The horizontal component of velocity remains unchanged (at 8sin 60º = 4√3 ms–1) throughout the motion of the packet since the gravitational force acts vertically and cannot affect the horizontal motion. The vertical component of velocity vvertical goes on increasing during the fall of the packet in accordance with the equation,

vvertical = uvertical + at where uvertical is the initial vertical component of velocity, a is the vertical acceleration and t is the time of flight of the packet.

Here uvertical = 8 cos 60º – 2 = 2 ms–1, a = g = 10 ms–2 and t = 2.968 s.

Therefore, vvertical = 2 + 10×2.968 = 31.68 ms–1.

v = √[(4√3)2 + (31.68)2] = 32.43 ms–1.

The direction of this velocity makes an angle θ with the horizontal (fig.) and is given by

tan θ = vvertical/ vhorizontal = 31.68/(4√3) = 4.572

Therefore θ = 77.66º

## Thursday, February 11, 2010

### Kinematics in Two Dimensions – Free Response Practice Question for AP Physics C

Today I’ll give you a free response practice question on Kinematics in Two Dimensions. Even though this question is meant for AP Physics C aspirants, those who prepare for AP Physics B exam also will find it useful. Here is the question:

A chopper at an air base is rising vertically up with a velocity of 2 ms–1. When it is at an altitude of 50 m, a packet is thrown down with velocity 8 ms–1 (with respect to the chopper), making an angle of 60º with the vertical. Neglecting air resistance and assuming that g = 10 ms–2, answer the following questions

(a) Draw a diagram to show the nature of the path AB followed by the packet from the moment it leaves the chopper to the moment it hits the ground, as seen by an observer on the ground. Is the path straight, circular or parabolic?.

(c) Calculate the time taken by the packet to reach the ground.

(d) Determine the horizontal distance traveled by the packet before hitting the ground.

(e) Determine the velocity (magnitude as well as direction) with which the packet hits the ground.

Try to answer this question.. You have 15 minutes at your disposal and can score up to 15 points for the right answer. Of course I’ll be back soon with a model answer for your benefit.

## Monday, February 1, 2010

### Kinematics in One Dimension –Multiple Choice Practice Questions for AP Physics B & C

All posts related to kinematics on this blog can be accessed either by clicking on the label ‘kinematics’ below this post or by performing a search using the search box at the top of this page. Today I give you a few multiple choice practice questions with solution:

(1) A stone projected vertically up is found to be at a height h at times 1 s and 3 s. Neglect air resistance and assume that g = 10 ms–2. The maximum height reached by the stone is

(a) 60 m

(b) 50 m

(c) 44 m

(d) 22 m

(e) 20 m

Evidently the stone reaches the same height h first during its upward trip and next during its downward trip. But you need not bother about this because everything is contained in the equation of motion, x x0 = v0t + ½ at2 (or, s = ut + ½ at2)

Considering the two cases given in the question, since a = g (taking upwrd as positive and therefore downward negative), we have

h = v0×1 – ½ ×10×12 = v0×3 – ½ ×10×32 from which

2v0 = 40 so that v0 = 20 ms–1

The naximum height hmax reached is given by by 0 = v02 – 2ghmax

[This is the equation, v2 = v02 + 2a(x x0) which can also be written as v2 = u2 + 2as]

Therefore, hmax = v02/2g = 202/(2×10) = 20 m

(2) A food packet is released from a helicopter ascending vertically up with uniform velocity of 2 ms–1. Neglect air resistance and assume that g = 10 ms–2. The velocity of the packet at the end of 0.2 s will be

(a) 5 ms–1

(b) 4 ms–1

(c) 2 ms–1

(d) 1 ms–1

(e) zero

The initial velocity v0 of the packet is 2 ms–1 and is directed upwards. The gravitational acceleration is directed downwards. The velocity v of the packet at time t is given by

v = v0gt.

[Note that we have taken the upward vector as positive and hence the downward vector as negative].

Substituting for v0, g and t, v = 2 – (10×0.2) = 0.

[After getting released, the packet moves upwards for 0.2 s with uniformly decreasing speed, momentarily comes to rest and then moves downwards with uniformly increasing speed].

(3) An astronaut on a planet having no atmosphere throws a stone of mass m vertically upwards with velocity 10 ms–1. It reaches maximum height of 10 m. If another stone of mass 2m is thrown upwards with velocity 20 ms–1, the maximum height reached will be

(a) 80 m

(b) 40 m

(c) 30 m

(d) 20 m

(e) 10 m

The maximum height (hmax) reached is given by

hmax = v02/2g

[Remember v2 = v02 + 2a(x x0) which can be written as v2 = v02 – 2ghmax for projection vertically apwards]

Substituting for hmax (= 10 m) and v0 (= 10 ms–1), the acceleration due to gravity (g) on the planet is obtained as 5 ms–2.

For a given velocity of projection, the height reached is independent of the mass of the stone.

When the velocity of projection is changed to 20 ms–1, the maximum height reached is 202/(2×5) = 40 m.

[You may note that the maximum height is directly proportional to the square of the velocity of projection. If the velocity of projection is doubled, the maximum height reached must be quadrupled as in the present problem].

The following question is specifically for AP Physics C aspirants. But AP Physics B aspirants too can ‘enjoy’ it.

(4) A ball thrown vertically up from the ground moves very close to an open window of a room in the first floor of a building. A student in the room sees the ball through the window, when the ball moves up as well as when it move down. The height of the window is 2 m and the ball is in view for a total time of 0.2 s. Take g = 10 ms–2. How high above the top of the window does the ball go (approximately)?

(a) 19 m

(b) 20 m

(c) 21 m

(d) 24 m

(e) 25 m

Since the total time for which the ball is in view is 0.2 s, the time taken by the ball to traverse the height of the window is 0.1 s. The average upward velocity (v) of the ball while moving past the window is 2/0.1 = 20 ms–1. We may take this to be very nearly equal to the velocity of the ball at the centre of the window.

Therefore, the maximum height reached by the ball (from the centre of the window) is given by

hmax = v2/2g = 202/(2×10) = 20 m.

[We have used the equation, v2 =u2 + 2as where a = g and s = hmax].

Therefore, the ball goes to a height of nearly 19 m above the top of the window [Option (a)].

[Since the motion of the ball is accelerated, the average velocity is not exactly equal to the velocity at the centre of the window. The answer we obtain is an approximate one.

If you wnt to calculate the exact height, you will proceed as follows:

If t is the time taken by the ball to fall from the topmost point of the trajectory to the top of the window, the maximum height hmax (measured from the top of the window) is given by

hmax = 0 + ½ gt2 = 5 t2 ………(i)

Also, (hmax + 2) = 0 + ½ g (t +0.1)2 = 5t2 + 5×0.01 + 5×2×0.1t

Or, hmax + 2 = 5 t2 + 0.05 + t ……….(ii)

Subtracting Eq (i) from Eq (ii) we obtain t = 1.95 s.

Therefore, hmax = 5 t2 = 5×(1.95)2 = 19.0125 m.