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.