Motion in a Plane — Page 2 | JEE Physics

Q11

A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be

A

20\sqrt 2

m

B 10 m

C

10\sqrt 2

m

D 20 m

Correct Answer

Option D

Solution

We know,

R = {{{u^2}{{\sin }2}\theta } \over g}

and

H = {{{u^2}{{\sin }^2}\theta } \over {2g}};

{H_{\max }}\,\,

is possible when

\theta = 90

^\circ

{H_{\max }} = {{{u^2}} \over {2g}} = 10 \Rightarrow {u^2} = 10g \times 2

As

R = {{{u^2}\sin 2\theta } \over g}

Range is maximum when projectile is thrown at an angle

45^\circ

.

\Rightarrow {R_{\max }} = {{{u^2}} \over g}

{R_{\max }} = {{10 \times g \times 2} \over g} = 20

meter

Q12

Ship A is sailing towards north-east with velocity

\mathop v\limits^ \to = 30\mathop i\limits^ \wedge + 50\mathop j\limits^ \wedge

km/hr where

\mathop i\limits^ \wedge

points east and

\mathop j\limits^ \wedge

, north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/hr. A will be at minimum distance from B in :

A 2.2 hrs

B 4.2 hrs

C 2.6 hrs

D 3.2 hrs

Correct Answer

Option C

Solution

Considering the initial position of ship A as origin, so the velocity and position of ship will be

{\overrightarrow v _A} = (30\widehat i + 50\widehat j)

and

{\overrightarrow r _A} = (0\widehat i + 0\widehat j)

Now, as given in the question, velocity and position of ship B will be,

{\overrightarrow v _B} = - 10\widehat i

and

{\overrightarrow r _B} = (80\widehat i + 150\widehat j)

Time after which the distance is minimum between A and B can be calculated as

t = {{|{{\overrightarrow r }_{BA}}.\,{{\overrightarrow v }_{BA}}|} \over {|{{\overrightarrow v }_{BA}}{|^2}}}

where,

{\overrightarrow r _{BA}} = {\overrightarrow r _B} - {\overrightarrow r _A} = 80\widehat i + 150\widehat j

and

{\overrightarrow v _{BA}} = - 10\widehat i - (30\widehat i + 50\widehat j)

= - 40\widehat i - 50\widehat j

\Rightarrow t = {{|(80\widehat i + 150\widehat j)\,.\,( - 40\widehat i - 50\widehat j)|} \over {| - 40\widehat i - 50\widehat j{|^2}}}

= {{3200 + 7500} \over {4100}} = {{10700} \over {4100}} = 2.6

h

Q13

Two particles are projected from the same point with the same speed u such that they have the same range R, but different maximum heights, h1 and h2. Which of the following is correct ?

A R2 = h1h2

B R2 = 16 h1h2

C R2 = 4 h1h2

D R2 = 2h1h2

Correct Answer

Option B

Solution

The range of two particles are same, that means angle of projections must be complementary to each other.

So one angle = $\theta$ and other one is = 90o - $\theta$ R =

{{{u^2}\sin 2\theta } \over g}

=

{{2{u^2}\sin \theta \cos \theta } \over g}

$\therefore$ R2 =

{{4{u^4}{{\sin }^2}\theta {{\cos }^2}\theta } \over {{g^2}}}

h1 =

{{{u^2}{{\sin }^2}\theta } \over {2g}}

h2 =

{{{u^2}{{\sin }^2}\left( {{{90}^o} - \theta } \right)} \over {2g}}

=

{{{u^2}{{\cos }^2}\theta } \over {2g}}

h1h2 =

{{{u^4}{{\sin }^2}\theta {{\cos }^2}\theta } \over {4{g^2}}}

$\Rightarrow$ h1h2 =

{{{R^2}} \over {16}}

$\Rightarrow$ R2 = 16 h1h2

Q14

A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t1 and t2 are the values of the time taken by it to hit the target in two possible ways, the product t1t2 is -

A

{{2R} \over g}

B

{R \over g}

C

{R \over {2g}}

D

{R \over {4g}}

Correct Answer

Option A

Solution

Range will be same for time t1 and t2, so angles of projection will be ‘ $\theta$ ’ & ‘90° – $\theta$ ’

{t_1} = {{2u\sin \theta } \over g}{t_2} = {{2u\sin \left( {{{90}^o} - \theta } \right)} \over g}

and

R = {{{u^2}\sin 2\theta } \over g}

{t_1}{t_2} = {{4{u^2}\sin \theta \cos \theta } \over {{g^2}}} = {2 \over g}\left[ {{{2{u^2}\sin \theta \cos \theta } \over g}} \right]

=

{{2R} \over g}

Q15

The stream of a river is flowing with a speed of 2km/h. A swimmer can swim at a speed of 4km/h. What should be the direction of the swimmer with respect to the flow of the river to cross the river straight ?

A 150°

B 120°

C 60°

D 90°

Correct Answer

Option B

Solution

Draw velocity diagram

\sin \theta = {{{v_r}} \over {{v_{sr}}}} = {1 \over 2}

\theta = {30^o}

$\phi$ = 90 + $\theta$ = 120°

Q16

A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60o with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, speed of the plane is :

A

{{\sqrt 3 } \over 2}

v

B

{{2v} \over {\sqrt 3 }}

C v

D

{v \over 2}

Correct Answer

Option D

Solution

AB = VP $\times$ t BC = Vt cos60o =

{{AB} \over {BC}}

{1 \over 2} = {{{V_P} \times t} \over {Vt}}

VP =

{V \over 2}

Q17

Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is -

A 1 : 16

B 1 : 8

C 1 : 2

D 1 : 4

Correct Answer

Option A

Solution

R = {{{u^2}\sin 2\theta } \over g}

A = \pi \,{R^2}

A \propto {R^2}

A \propto {u^4}

{{{A_1}} \over {{A_2}}} = {{u_1^4} \over {u_2^4}} = {\left[ {{1 \over 2}} \right]^4} = {1 \over {16}}

Q18

The position co-ordinates of a particle moving in a 3-D coordinate system is given by x = a cos

\omega

t y = a sin

\omega

t and z = a

\omega

t The speed of the particle is :

A

\sqrt 2 \,a\omega

B

a\omega

C

\sqrt 3 \,a\omega

D 2a

\omega

Correct Answer

Option A

Solution

Given that, x = a cos $\omega$ t y = a sin $\omega$ t z = a $\omega$ t Velocity in x-direction, Vx =

{{dx} \over {dt}} = - a\omega \sin \omega t

Velocity in y-direction, Vy =

{{dy} \over {dt}}

= a $\omega$ cos $\omega$ t Velocity in z-direction, Vz =

{{dz} \over {dt}}

= a $\omega$ Net velocity,

\overrightarrow V

= Vx

\widehat i

+ Vy

\widehat j

+ Vz

\widehat k

Speed =

\left| {\overrightarrow V } \right| = \sqrt {V_x^2 + V_y^2 + V_z^2}

= \sqrt {{a^2}{\omega ^2}{{\sin }^2}\omega t + {a^2}{\omega ^2}{{\cos }^2}\omega t + {a^2}{\omega ^2}}

= \sqrt {{a^2}{\omega ^2}\left( {{{\sin }^2}\omega t + {{\cos }^2}\omega t} \right) + {a^2}{\omega ^2}}

= \sqrt {2{a^2}{\omega ^2}}

= \sqrt 2 a\omega

Q19

When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed v, he sees that rain drops are coming at an angle 60° from the horizontal. On further increasing the speed of the car to (1 +

\beta

)v, this angle changes to 45o. The value of

\beta

is close to :

A 0.50

B 0.73

C 0.37

D 0.41

Correct Answer

Option B

Solution

tan 60o =

{{{V_r}} \over V}

.....(1) tan 45o =

{{{V_r}} \over {\left( {1 + \beta } \right)V}}

.....(2) From (i) and (ii), we get

{{\sqrt 3 } \over 1} = {{{1 \over V}} \over {{1 \over {\left( {1 + \beta } \right)V}}}}

$\Rightarrow$

{\sqrt 3 = \left( {1 + \beta } \right)}

$\Rightarrow$ $\beta$ = 0.732

Q20

Starting from the origin at time t = 0, with initial velocity 5

\widehat j

ms-1 , a particle moves in the x-y plane with a constant acceleration of

\left( {10\widehat i + 4\widehat j} \right)

ms-2. At time t, its coordinates are (20 m, y0 m). The values of t and y0 are, respectively:

A 5s and 25 m

B 2s and 18 m

C 2s and 24 m

D 4s and 52 m

Correct Answer

Option B

Solution

y = {u_y}t + {1 \over 2}{a_y}{t^2}

y = 5t + {1 \over 2}(4){t^2}

y = 5t + 2{t^2}

and

x = 0(t) + {1 \over 2}(10)({t^2}) = 20

t = 2s

\Rightarrow y = 10 + 8 = 18m