*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 *at*^{2} 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 = *40*T and H = *15*T *– 1.25*T*^{2}^{}

*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*_{0x}*T *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*_{0x}^{2} + *v*_{0y}^{2}) = √(40^{2} + 30^{2}) = 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 –2*t* (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 =* 30*t *and *y* = 40*t** _{ }*– 5

*t*

^{2}. 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 – 10*t*

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 =* 30*t *and* dx/dt = v*_{x}* = **v*_{0x} = 30 ms^{–1}. [Since *x = *30*t*, 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*_{0x}^{2} + *v*_{0y}^{2}) = √(30^{2} + 40^{2}) = 50 ms^{–1}.

**(2) **The displacement** **of a particle at time *t *has *x-*component 3 + *at*^{3} and *y*-component 4 + *bt*^{3}. The speed of the particle at time *t *is

(a) 3*t*^{2}√(*a*^{ 2} + *b*^{2})

(b) 3*t*^{2}(*a*^{ 2} + *b*^{2})

(c) 3*t*^{}√(*a*^{ 2} + *b*^{2})

(d) 3*t*^{}(*a*^{ 2} + *b*^{2})

(e)** ***t*^{2}√(*a*^{ 2} + *b*^{2})** **

The *x-*component of velocity (*v*_{x})* *is the time derivative of the *x-*component of displacement: *v*_{x} = 3* at*^{2}

Similarly, *y-*component of velocity, *v*_{y} = 3* bt*^{2}

The magnitude (*v*) of the velocity at time *t* = √(*v*_{x}^{2} + *v*_{y}^{2})

Therefore, *v *=√[(3* at*^{2 })^{2} + (3* bt*^{2})^{2}] = 3*t*^{2}√(*a*^{ 2} + *b*^{2})

This is the speed of the particle at time *t.*

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