Motion in a Plane Questions — JEE Physics

Q1

A projectile can have the same range 'R' for two angles of projection. If T1 and T2 be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

A R

B

{1 \over R}

C

{1 \over {{R^2}}}

D

{R^2}

Correct Answer

Option A

Solution

Range is same for angle of projection

\theta ,

and

{90^ \circ } - \theta

{T_1} = {{2u\sin \theta } \over g},\,\,{T_2} = {{2u\cos \theta } \over g}

{T_1}{T_2} =

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

=

{2 \over g} \times \left( {{{{u^2}\sin 2\theta } \over g}} \right)

=

{{2R} \over g}

(as

R =

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

) Hence,

{T_1}{T_2}

is proportional to

R.

Q2

A particle starts from the origin at t = 0 with an initial velocity of 3.0

\widehat i

m/s and moves in the x-y plane with a constant acceleration

\left( {6\widehat i + 4\widehat j} \right)

m/s2 . The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D meters. The value of D is :-

A 40

B 32

C 50

D 60

Correct Answer

Option D

Solution

\overrightarrow u

= 3.0

\widehat i

\overrightarrow a

=

\left( {6\widehat i + 4\widehat j} \right)

\overrightarrow S = \overrightarrow u t + {1 \over 2}\overrightarrow a {t^2}

x = 3t +

{1 \over 2}6{t^2}

= 3t + 3t2 .....(1) y =

{1 \over 2} \times 4 \times {t^2}

= 32 $\Rightarrow$ t = 4 s .... (2) x = 3 × 4 + 3 × 42 = 12 + 48 = 60 m

Q3

A projectile is given an initial velocity of

\left( {\widehat i + 2\widehat j} \right)

m/s, where

{\widehat i}

is along the ground and

{\widehat j}

is along the vertical. If g = 10 m/s2, the equation of its trajectory is:

A y = x - 5x2

B y = 2x - 5x2

C 4y = 2x - 5x2

D 4y = 2x - 25x2

Correct Answer

Option B

Solution

\overrightarrow u = \widehat i + 2\widehat j = {u_x}\widehat i + {u_y}\widehat j

\Rightarrow u\cos \theta = 1,u\sin \theta = 2

Also

x = {u_x}t

and

y = {u_y}t - {1 \over 2}g{t^2}

$\Rightarrow$

y = x\tan \theta - {1 \over 2}{{g{x^2}} \over {u_x^2}}

$\therefore$

y = 2x - {1 \over 2}g{x^2} = 2x - 5{x^2}

Q4

The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance upto which he can throw the same ball is :

A 136 m

B 272 m

C 68 m

D 192 m

Correct Answer

Option B

Solution

For vertical throw,

\begin{aligned} & h=\frac{v^{2}}{2 g} \\\\ & v=\sqrt{2 g h}=\sqrt{2 g \times 136} \quad...(1) \end{aligned}

For max range, $\theta=45^{\circ}$ $R_{\max }=\dfrac{v^{2}}{g} \quad...(2)$ From (1) and (2)

\begin{aligned} R_{\max } & =\frac{v^{2}}{g}=\frac{2 g \times 136}{g} \\\\ & =272 \mathrm{~m} \end{aligned}

Q5

A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of

30^\circ

with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground?

\left[ {g = 10m/{s^2},\sin 30^\circ = {1 \over 2},\cos 30^\circ = {{\sqrt 3 } \over 2}} \right]

A 5.20 m

B 4.33 m

C 2.60 m

D 8.66 m

Correct Answer

Option D

Solution

From the figure it is clear that maximum horizontal range

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

= {{{{\left( {10} \right)}^2}\sin \left( {2 \times {{30}^ \circ }} \right)} \over {10}}

= 5\sqrt 3

= 8.66 m

Q6

A ball is thrown from a point with a speed ν0 at an angle of projection θ. From the same point and at the same instant person starts running with a constant speed

{{{v_0}} \over 2}

to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection θ?

A No

B Yes,

30^\circ

C Yes,

60^\circ

D Yes,

45^\circ

Correct Answer

Option C

Solution

Yes, the person can catch the ball when horizontal velocity is equal to the horizontal component of ball's velocity, the motion of ball will be only in vertical direction with respect to person for that,

{{{v_0}} \over 2} = {v_0}\cos \theta \,\,\,\,

or

\cos \theta = {1 \over 2}

\Rightarrow \cos \theta = \cos 60^\circ

\Rightarrow \theta = 60^\circ

Q7

A particle is moving eastwards with a velocity of 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is

A

{1 \over 2}m{s^{ - 2}}

towards north

B

{1 \over {\sqrt 2 }}m{s^{ - 2}}

towards north-east

C

{1 \over {\sqrt 2 }}m{s^{ - 2}}

towards north-west

D zero

Correct Answer

Option C

Solution

Average acceleration

= {{change\,\,in\,\,velocioty} \over {time\,\,{\mathop{\rm \int}} erval}}

= {{\Delta \overrightarrow v } \over t}

{\overrightarrow v _1} = +5\widehat i

, towards east direction.

\overrightarrow {{v_2}} = +5\widehat j

, towards north direction

\Delta \overrightarrow v = \overrightarrow {{v_2}} - \overrightarrow {{v_1}}

=

5\widehat i - 5\widehat j

$\therefore$

\,\,\,\,\overrightarrow a = {{5\widehat j - 5\widehat i} \over {10}} = {{\widehat j - \widehat i} \over 2}

$\therefore$

\,\,\,\, a = {{\sqrt {{1^2} + {{\left( { - 1} \right)}^2}} } \over 2} = {{\sqrt 2 } \over 2} = {1 \over {\sqrt 2 }}m{s^{ - 2}}

\tan \theta = {{{v_2}} \over {{v_1}}} = {5 \over 5} = 1

$\therefore$

\,\,\,\,\theta = {45^ \circ }

Therefore the direction is North-west.

Q8

A particle has an initial velocity

3\widehat i + 4\widehat j

and an acceleration of

0.4\widehat i + 0.3\widehat j

. Its speed after 10 s is:

A

7\sqrt 2

units

B 7 units

C 8.5 units

D 10 units

Correct Answer

Option A

Solution

Given

\overrightarrow u = 3\widehat i + 4\widehat j,\,\,\overrightarrow a = 0.4\widehat i + 0.3\widehat j,\,\,t = 10s

\overrightarrow v = \overrightarrow u + \overrightarrow a t

= 3\widehat i + 4\widehat j + \left( {0.4\widehat i + 0.3\widehat j} \right) \times 10

= 7\widehat i + 7\widehat j

We know speed is equal to magnitude of velocity. $\therefore$

\left| {\overrightarrow v } \right| = \sqrt {{7^2} + {7^2}} = 7\sqrt 2 \,\,\,

units

Q9

A particle is moving with velocity

\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)

, where K is a constant. The general equation for its path is

A y = x2 + constant

B y2 = x + constant

C xy = constant

D y2 = x2 + constant

Correct Answer

Option D

Solution

\overrightarrow v = k\left( {y\widehat i + x\widehat j} \right)

........(1) Also

\overrightarrow v = {v_x}\widehat i + {v_y}\widehat j

\overrightarrow v = {{dx} \over {dt}}\widehat i + {{dy} \over {dt}}\widehat j

........(2) Equating (1) and (2), we get

{{dx} \over {dt}} = ky\,\,\,\,\,\,

.......(3) and

\,\,\,\,\,{{dy} \over {dt}} = kx

......(4) Dividing (3) and (4), we get

{{dy} \over {dx}} = {x \over y}

\Rightarrow ydy = xdx

Integrating both sides of above equation, we get

\int {ydy} = \int {xdx}

\Rightarrow {y^2} = {x^2} +

constant

Q10

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is :

A

\pi {{{v^4}} \over {{g^2}}}

B

{\pi \over 2}{{{v^4}} \over {{g^2}}}

C

\pi {{{v^2}} \over {{g^2}}}

D

\pi {{{v^2}} \over g}

Correct Answer

Option A

Solution

Maximum range of water coming out of fountain,

{R_{\max }} = {{{v^2}\sin 2\theta } \over g} = {{{v^2}\sin {{90}^ \circ }} \over g} = {{{v^2}} \over g}

Total area around fountain,

A = \pi R_{\max }^2\,\, = \,\,\pi {{{v^4}} \over {{g^2}}}