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## Tuesday, September 2, 2008

### 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 27th 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,

Tf = 2 vy/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 = Tf]

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 = v0xT where v0x 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 = v0y(T/2) – ½ g(T/2)2]

The velocity of projection is √(v0x2 + v0y2) = √(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 (vy) of velocity at the instant t during the flight of the projectile is given by

vy = dy/dt = 40 – 10t

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

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

The initial vertical component of velocity (v0y) 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 = vx = v0x = 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 √(v0x2 + v0y2) = √(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 (vx) is the time derivative of the x-component of displacement: vx = 3 at2

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

The magnitude (v) of the velocity at time t = √(vx2 + vy2)

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

This is the speed of the particle at time t.