Practice Start Test

Motion in a Plane

NEET Physics · 49 questions · Page 2 of 5 · Click an option or "Show Solution" to reveal answer

Q11
A particle moving with velocity V→\overrightarrow V is acted by three forces shown by the vector triangle PQR. The velocity of the particle will :
A remain constant
B increase
C decrease
D change according to the smallest force QR→\overrightarrow {QR}
Correct Answer
Option A
Solution

From the given diagram, the forces are forming closed loop in same order. ∴\therefore F net =

0→\overrightarrow 0

⇒\Rightarrow

mdvdtm{{dv} \over {dt}}

=

0→\overrightarrow 0

⇒\Rightarrow

v→\overrightarrow v

= constant.

Q12
When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60 o with horizontal, it can travel a distance x 1 along the plane. But when the inclination is decreased to 30 o and the same object is shot with the same velocity, it can travel x 2 distance. Then x 1 : x 2 will be :
A 1 : 2\sqrt 2
B 2\sqrt 2 : 1
C 1 : 3\sqrt 3
D 1 : 23\sqrt 3
Correct Answer
Option C
Solution

Assume initial velocity = u In first case, Moved distance, (x 1 ) =

u22gsin⁡60o{{{u^2}} \over {2g\sin {{60}^o}}}

In second case, Moved distance, (x 2 ) =

u22gsin⁡30o{{{u^2}} \over {2g\sin {{30}^o}}}

∴\therefore

x1x2=sin⁡30osin⁡60o{{{x_1}} \over {{x_2}}} = {{\sin {{30}^o}} \over {\sin {{60}^o}}}

=

1×22×3{{1 \times 2} \over {2 \times \sqrt 3 }}

=

13{1 \over {\sqrt 3 }}
Q13
The x and y coordinates of the particle at any time are x = 5t −- 2t 2 and y = 10t respectively, where x and y are in metres and t in seconds. The acceleration of the particle at t = 2 s is
A 5 m s −-2
B −-4 m s −-2
C −-8 m s −-2
D 0
Correct Answer
Option B
Solution

x = 5t −- 2t 2 and y = 10t

dxdt{{dx} \over {dt}}

= 5 - 4t,

dydt{{dy} \over {dt}}

= 10 ∴\therefore v x = 5 - 4t, v y = 10

dvxdt{{d{v_x}} \over {dt}}

= - 4,

dvydt{{d{v_y}} \over {dt}}

= 0 ∴\therefore

ax{a_x}

= - 4,

ay{a_y}

= 0

a→=axi^+ayj^\overrightarrow a = {a_x}\widehat i + {a_y}\widehat j

=

−4j^- 4\widehat j

m/s 2 ∴\therefore Acceleration of the particle at t = 2 s is - 4 m/s 2

Q14
A particle moves so that its position vector is given by r→=cos⁡ωt x^+sin⁡ ωt y^,\overrightarrow r = \cos \omega t\,\widehat x + \sin \,\omega t\,\widehat y, where ω\omega is a constant. Which of the following is true?
A Velocity is perpendicular to r→\overrightarrow r and acceleration is directed towards the origin.
B Velocity is perpendicular to r→\overrightarrow r and acceleration is directed away from the origin.
C Velocity and acceleration both are perpendicular to r→\overrightarrow r
D Velocity and acceleration both are parallel to r→\overrightarrow r
Correct Answer
Option A
Solution
r→=cos⁡ωt x^+sin⁡ ωt y^,\overrightarrow r = \cos \omega t\,\widehat x + \sin \,\omega t\,\widehat y,

∴\therefore

v→=−ωsin⁡ωtx^+ωcos⁡ωty^\overrightarrow v = - \omega \sin \omega t\widehat x + \omega \cos \omega t\widehat y

and

a→=−ω2cos⁡ωtx^−ω2sin⁡ωty^\overrightarrow a = - {\omega ^2}\cos \omega t\widehat x - {\omega ^2}\sin \omega t\widehat y

=

−ω2r→- {\omega ^2}\overrightarrow r
r→.v→\overrightarrow r .\overrightarrow v

= 0 ⇒\Rightarrow

r→⊥v→\overrightarrow r \bot \overrightarrow v

As position vector

(r→)\left( {\overrightarrow r } \right)

is directly away from the origin, so, acceleration (

−ω2r→- {\omega ^2}\overrightarrow r

) is directed towards the origin.

Q15
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is
A 45 o
B 180 o
C 0 o
D 90 o
Correct Answer
Option D
Solution

Let the two vectors are

A→\overrightarrow A

and

B→.\overrightarrow B.

Given that,

∣A→+B→∣=∣A→−B→∣\left| {\overrightarrow A + \overrightarrow B } \right| = \left| {\overrightarrow A - \overrightarrow B } \right|

∴\therefore

A2+B2+2ABcos⁡θ\sqrt {{A^2} + {B^2} + 2AB\cos \theta }

=

A2+B2−2ABcos⁡θ\sqrt {{A^2} + {B^2} - 2AB\cos \theta }

⇒\Rightarrow

4ABcos⁡θ{4AB\cos \theta }

= 0 ∵\because

4AB{4AB}

≠\ne 0 ∴\therefore

cos⁡θ{\cos \theta }

= 0 ⇒\Rightarrow θ\theta = 90 o

Q16
The positions vector of a particle R→\overrightarrow R as a function of time is given by R→\overrightarrow R = 4sin(2π\pi t)i^\widehat i + 4cos(2π\pi t)j^\widehat j. Where R is in meters, t is in seconds and i^\widehat i and j^\widehat j denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of particle?
A Magnitude of the velocity of particle is 8 meter/second.
B Path of the particle is a circle of radius 4 meter.
C Acceleration vector is along −-R→\overrightarrow R .
D Magnitude of acceleration vector is v2R{{{v^2}} \over R} where v is the velocity of particle.
Correct Answer
Option A
Solution

Here,

R→=4sin⁡(2πt)i^+4cos⁡(2πt)j^\overrightarrow R = 4\sin \left( {2\pi t} \right)\widehat i + 4\cos \left( {2\pi t} \right)\widehat j

The velocity of the particle is

v→=dR→dt=ddt[4sin⁡(2πt)i^+4cos⁡(2πt)j^]\overrightarrow v = {{d\overrightarrow R } \over {dt}} = {d \over {dt}}\left[ {4\sin \left( {2\pi t} \right)\widehat i + 4\cos \left( {2\pi t} \right)\widehat j} \right]
=8πcos(2πt)i^−8πsin(2πt)j^= 8\pi cos\left( {2\pi t} \right)\widehat i - 8\pi sin\left( {2\pi t} \right)\widehat j

Its magnitude is

∣v→∣=(8πcos⁡(2πt))2+(−8πsin⁡(2πt))2\left| {\overrightarrow v } \right| = \sqrt {{{\left( {8\pi \cos \left( {2\pi t} \right)} \right)}^2} + {{\left( { - 8\pi \sin \left( {2\pi t} \right)} \right)}^2}}
=64π2cos⁡2(2πt)+64π2sin⁡2(2πt)= \sqrt {64{\pi ^2}{{\cos }^2}\left( {2\pi t} \right) + 64{\pi ^2}{{\sin }^2}\left( {2\pi t} \right)}
=64π2[cos⁡2(2πt)+sin⁡2(2πt)]= \sqrt {64{\pi ^2}\left[ {{{\cos }^2}\left( {2\pi t} \right) + {{\sin }^2}\left( {2\pi t} \right)} \right]}
=64π2= \sqrt {64{\pi ^2}}

(as sin 2 θ\theta + cos 2 θ\theta = 1) = 8π\pi m/s

Q17
If vectors A→=cos⁡ωti^+sin⁡ωtj^\overrightarrow A = \cos \omega t\widehat i + \sin \omega t\widehat j and B→=cos⁡ωt2i^+sin⁡ωt2j^\overrightarrow B = \cos {{\omega t} \over 2}\widehat i + \sin {{\omega t} \over 2}\widehat j are functions of time, then the value of t at which they are orthogonal to each other is
A t=πωt = {\pi \over \omega }
B t == 0
C t=π4ωt = {\pi \over {4\omega }}
D t=π2ωt = {\pi \over {2\omega }}
Correct Answer
Option A
Solution

Two vectors

A‾\overline A

and

B‾\overline B

are orthogonal to each other, if their scalar product is zero i.e.

A‾\overline A

.

B‾\overline B

= 0. Here,

A‾=cos⁡ωti^+sin⁡ωtj^\overline A = \cos \omega t\widehat i + \sin \omega t\widehat j

and

B‾=cos⁡ωt2i^+sin⁡ωt2j^\overline B = \cos {{\omega t} \over 2}\widehat i + \sin {{\omega t} \over 2}\widehat j
∴A‾.B‾=(cos⁡ωti^+sin⁡ωtj^)(cos⁡ωt2i^+sin⁡ωt2j^)\therefore \overline A .\overline B = \left( {\cos \omega t\widehat i + \sin \omega t\widehat j} \right)\left( {\cos {{\omega t} \over 2}\widehat i + \sin {{\omega t} \over 2}\widehat j} \right)

=

cos⁡ωtcos⁡ωt2+sin⁡ωtsin⁡ωt2\cos \omega t\cos {{\omega t} \over 2} + \sin \omega t\sin {{\omega t} \over 2}
(∵i^.i^=j^.j^=1 and i^.j^=j^.i^=0)\left( \because {\widehat i.\widehat i = \widehat j.\widehat j = 1\,and\,\widehat i.\widehat j = \widehat j.\widehat i = 0} \right)

=

cos⁡(ωt−ωt2)\cos \left( {\omega t - {{\omega t} \over 2}} \right)

(

∵cos⁡(A−B)=cosAcosB+sinAsinB\because \cos (A - B) = cosAcosB + sinAsinB

) But

A‾.B‾=0\overline A .\overline B = 0

(as

A‾\overline A

and

B‾\overline B

are orthogonal to each other)

∴cos⁡(ωt−ωt2)=0\therefore \cos \left( {\omega t - {{\omega t} \over 2}} \right) = 0
cos⁡(ωt−ωt2)=cos⁡π2  or ωt−ωt2=π2\cos \left( {\omega t - {{\omega t} \over 2}} \right) = \cos {\pi \over 2}\,\,or\,\omega t - {{\omega t} \over 2} = {\pi \over 2}
ωt2=π2 or t=πω{{\omega t} \over 2} = {\pi \over 2}\,or\,t = {\pi \over \omega }
Q18
A ship A is moving Westwards with a speed of 10 km h −-1 and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h −-1 . The time after which the distance between them becomes shortest, is
A 525\sqrt 2 h
B 10210\sqrt 2 h
C 0 h
D 5 h
Correct Answer
Option D
Solution
V→A=10(−i^){\overrightarrow V _A} = 10\left( { - \widehat i} \right)
V→B=10(j^){\overrightarrow V _B} = 10\left( {\widehat j} \right)
V→BA=10j^+10i^=102 km/h{\overrightarrow V _{BA}} = {\rm{10}}\widehat {\rm{j}}{\rm{ + 10}}\widehat i{\rm{ = 10}}\sqrt 2 \,{\rm{km/h }}

Distance OB = 100 cos 45° = 50

2\sqrt 2

km Time taken to reach the shortest distance between A and B =

OBVBA→=502102=5h{{OB} \over {\overrightarrow {{V_{BA}}} }} = {{50\sqrt 2 } \over {10\sqrt 2 }} = 5h
Q19
A projectile is fired from the surface of the earth with a velocity of 5 m s −-1 and angle θ\theta with the horizontal. Another projectile fired from another planet with a velocity of 3 m s −-1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in m s −-2 ) is (Given g = 9.8 m s −-2 )
A 3.5
B 5.9
C 16.3
D 110.8
Correct Answer
Option A
Solution

The equation of trajectory is

y=xtan⁡θ−gx22u2cos⁡2θy = x\tan \theta - {{g{x^2}} \over {2{u^2}{{\cos }^2}\theta }}

where θ\theta is the angle of projection and u is the velocity with which projectile is projected.

For equal trajectories and for same angles of projection,

gu2{g \over {{u^2}}}

= constant According to the question,

9.852=g′32{{9.8} \over {{5^2}}} = {{g'} \over {{3^2}}}

where g' is acceleration due to gravity on the planet.

g′=9.8×925=3.5 ms−2g' = {{9.8 \times 9} \over {25}} = 3.5\,m{s^{ - 2}}
Q20
A particle is moving such that is position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at time t = 2 s and (13 m, 14 m) at time t = 5 s. Average velocity vector (v→av)\left( {{{\overrightarrow v }_{av}}} \right) from t = 0 to t = 5 s is
A 15(13i^+14j^){1 \over 5}\left( {13\widehat i + 14\widehat j} \right)
B 73(i^+j^){7 \over 3}\left( {\widehat i + \widehat j} \right)
C 2(i^+j^)2\left( {\widehat i + \widehat j} \right)
D 115(i^+j^){11 \over 5}\left( {\widehat i + \widehat j} \right)
Correct Answer
Option D
Solution
vav→=Δr→ (displacement)Δt (time taken)\overrightarrow {{v_{av}}} = {{\Delta \overrightarrow r \,(displacement)} \over {\Delta t\,(time\,taken)}}
=(13−2)i^+(14−3)j^5−0=115(i+j)= {{\left( {13 - 2} \right)\widehat i + \left( {14 - 3} \right)\widehat j} \over {5 - 0}} = {{11} \over 5}\left( {i + j} \right)
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