Practice Start Test

Motion in a Plane

NEET Physics · 49 questions · Page 3 of 5 · Click an option or "Show Solution" to reveal answer

Q21
Vectors A→,B→\overrightarrow A ,\overrightarrow B and C→\overrightarrow C are such that A→.B→=0\overrightarrow A .\overrightarrow B = 0 and A→.C→=0\overrightarrow A .\overrightarrow C = 0. Then the vector parallel to A→\overrightarrow A is
A A→×B→\overrightarrow A \times \overrightarrow B
B B→+C→\overrightarrow B + \overrightarrow C
C B→×C→\overrightarrow B \times \overrightarrow C
D B→\overrightarrow B and C→\overrightarrow C
Correct Answer
Option C
Solution

Vector triple product of three vectors

A→\overrightarrow A

,

B→\overrightarrow B

and

C→\overrightarrow C

is

A→×(B→×C→)=(A→.C→)B→−(A→.B→)C→\overrightarrow A \times \left( {\overrightarrow B \times \overrightarrow C } \right) = \left( {\overrightarrow A .\overrightarrow C } \right)\overrightarrow B - \left( {\overrightarrow A .\overrightarrow B } \right)\overrightarrow C

Given:

A→.B→=0,A→.C→=0\overrightarrow A .\overrightarrow B = 0,\overrightarrow A .\overrightarrow C = 0

∴\therefore

A→×(B→×C→)=0\overrightarrow A \times \left( {\overrightarrow B \times \overrightarrow C } \right) = 0

Thus the vector

A→\overrightarrow A

is parallel to vector

B→×C→{\overrightarrow B \times \overrightarrow C }

.

Q22
The velocity of a projectile at the initial point A is (2i^+3j^)\left( {2\widehat i + 3\widehat j} \right) m/s. It's velocity (in m/s) at point B is
A 2i^−3j^2\widehat i - 3\widehat j
B 2i^+3j^2\widehat i + 3\widehat j
C −-2i^−3j^2\widehat i - 3\widehat j
D −-2i^+3j^2\widehat i + 3\widehat j
Correct Answer
Option A
Solution

At point B X component of velocity remains unchanged while Y component reverses its direction.

∴\therefore The velocity of the projectile at point B is

2i^−3j^2\widehat i - 3\widehat j

m/s.

Q23
A particle has initial velocity (2i→+3j→)\left( {2\overrightarrow i + 3\overrightarrow j } \right) and acceleration (0.3i→+0.2j→)\left( {0.3\overrightarrow i + 0.2\overrightarrow j } \right). The magnitude of velocity after 10 seconds will be
A 929\sqrt 2
B 525\sqrt 2
C 5 units
D 9 units
Correct Answer
Option B
Solution
P→\overrightarrow P

= vector sum =

A→+B→\overrightarrow A + \overrightarrow B
Q→\overrightarrow Q

= Vector differences =

A→−B→\overrightarrow A - \overrightarrow B

Since

P→\overrightarrow P

and

Q→\overrightarrow Q

are perpendicular ∴\therefore

P→.Q→=0⇒(A→+B→).(A→−B→)=0\overrightarrow P .\overrightarrow Q = 0 \Rightarrow \left( {\overrightarrow A + \overrightarrow B } \right).\left( {\overrightarrow A - \overrightarrow B } \right) = 0
⇒A2=B2=∣A→∣=∣B→∣\Rightarrow {A^2} = {B^2} = \left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right|
Q24
The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is
A θ\theta = tan −-1 (14)\left( {{1 \over 4}} \right)
B θ\theta = tan −-1 (4)
C θ\theta = tan −-1 (2)
D θ\theta = 45 o
Correct Answer
Option B
Solution

Horizontal range

R=u2sin⁡2θgR = {{{u^2}\sin 2\theta } \over g}

......(1) Maximum height

H=u2sin⁡2θ2gH = {{{u^2}{{\sin }^2}\theta } \over {2g}}

.....(2) According to the problem R = H

u2sin⁡2θg{{{u^2}\sin 2\theta } \over g}

=

u2sin⁡2θ2g{{{u^2}{{\sin }^2}\theta } \over {2g}}
⇒2sin⁡θcos⁡θ=sin⁡2θ2\Rightarrow 2\sin \theta \cos \theta = {{{{\sin }^2}\theta } \over 2}
2cos⁡θ=sin⁡θ22\cos \theta = {{\sin \theta } \over 2}
⇒cot⁡θ=14\Rightarrow \cot \theta = {1 \over 4}
⇒tan⁡θ=4\Rightarrow \tan \theta = 4
⇒θ=tan⁡−1(4)\Rightarrow \theta = {\tan ^{ - 1}}(4)
Q25
A projectile is fired at an angle of 45 o with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is
A 45 o
B 60 o
C tan −-1 (12)\left( {{1 \over 2}} \right)
D tan −-1 (32)\left( {{{\sqrt 3 } \over 2}} \right)
Correct Answer
Option C
Solution

Let ϕ\phi be elevation angle of the projectile at its highest point as seen from the point of projection O and θ\theta be angle of projection with the horizontal.

From figure, tanϕ\phi =

HR/2{H \over {R/2}}

....(1) In case of projectile motion Maximum height H =

u2sin⁡2θ2g{{{u^2}{{\sin }^2}\theta } \over {2g}}

Horizontal range, R =

u2sin⁡θg{{{u^2}\sin \theta } \over g}

Substituting these values of H and R in (1), we get

tan⁡ϕ=u2sin⁡2θ2gu2sin⁡2θ2g\tan \phi = {{{{{u^2}{{\sin }^2}\theta } \over {2g}}} \over {{{{u^2}\sin 2\theta } \over {2g}}}}
tan⁡ϕ=sin⁡2θsin⁡2θ=sin⁡2θ2sin⁡θcos⁡θ\tan \phi = {{{{\sin }^2}\theta } \over {\sin 2\theta }} = {{{{\sin }^2}\theta } \over {2\sin \theta \cos \theta }}

=

12tan⁡θ{1 \over 2}\tan \theta
tan⁡ϕ=12tan⁡45∘=12\tan \phi = {1 \over 2}\tan 45^\circ = {1 \over 2}

Here, θ\theta = 45° ∴\therefore

tan⁡ϕ=12tan⁡45∘=12\tan \phi = {1 \over 2}\tan 45^\circ = {1 \over 2}
(∵tan⁡45∘=1)\left(\because {\tan 45^\circ = 1} \right)
ϕ=tan⁡−1(12)\phi = {\tan ^{ - 1}}\left( {{1 \over 2}} \right)
Q26
A body is moving with velocity 30 m/s towards east . After 10 seconds its velocity becomes 40 m/s towards north. The average acceleration of the body is
A 1 m/s 2
B 7 m/s 2
C 7\sqrt 7 m/s 2
D 5 m/s 2
Correct Answer
Option D
Solution

Velocity towards east direction,

v1→=30i^\overrightarrow {{v_1}} = 30\widehat i

m/s Velocity towards north direction,

v2→=40j^\overrightarrow {{v_2}} = 40\widehat j

m/s Change in velocity,

Δv→=v→2−v→1\Delta \overrightarrow v = {\overrightarrow v _2} - {\overrightarrow v _1}

=

(40j^−30i^)\left( {40\widehat j - 30\widehat i} \right)
∴∣Δv→∣=∣40j^−30i^∣=50 m/s\therefore \left| {\Delta \overrightarrow v } \right| = \left| {40\widehat j - 30\widehat i} \right| = 50\,m/s

Average acceleration,

a→av{\overrightarrow a _{av}}

=

change in velocityTime interval{{change\,in\,velocity} \over {Time\, interval}}
a→av=v→2−v→1Δt=Δv→Δt{\overrightarrow a _{av}} = {{{{\overrightarrow v }_2} - {{\overrightarrow v }_1}} \over {\Delta t}} = {{\Delta \overrightarrow v } \over {\Delta t}}
∣a→av∣=∣Δv→∣Δt=50 m/s10 s=5 m/s2\left| {{{\overrightarrow a }_{av}}} \right| = {{\left| {\Delta \overrightarrow v } \right|} \over {\Delta t}} = {{50\,m/s} \over {10\,s}} = 5\,m/{s^2}
Q27
A missile is fired for maximum range with an initial velocity of 20 m/s. If g = 10 m/s 2 , the range of the missile is
A 40 m
B 50 m
C 60 m
D 20 m
Correct Answer
Option A
Solution

For maximum range, the angle of projection, θ\theta = 45°.

∴R=u2sin⁡2θg\therefore R = {{{u^2}\sin 2\theta } \over g}
=(20)2sin⁡(2×45∘)10= {{{{\left( {20} \right)}^2}\sin \left( {2 \times {{45}^\circ }} \right)} \over {10}}
=400×110=40 m= {{400 \times 1} \over {10}} = 40\,m
Q28
A particle moves in a circle of radius 5 cm with constant speed and time period 0.2π\pi s. The acceleration of the particle is
A 15 m/s 2
B 25 m/s 2
C 36 m/s 2
D 5 m/s 2
Correct Answer
Option D
Solution

Centripetal acceleration

ac=ω2r{a_c} = {\omega ^2}r
=(2πT)2r= {\left( {{{2\pi } \over T}} \right)^2}r
=(2π0.2π)2×5×10−2=5= {\left( {{{2\pi } \over {0.2\pi }}} \right)^2} \times 5 \times {10^{ - 2}} = 5

m/s 2

Q29
The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is
A 60 o
B 15 o
C 30 o
D 45 o
Correct Answer
Option A
Solution

Let v be velocity of a projectile at maximum height H. v = ucosθ\theta According to given problem,

v=u2v = {u \over 2}

∴\therefore

u2=ucos⁡θ⇒cos⁡θ=12⇒θ=60∘{u \over 2} = u\cos \theta \Rightarrow \cos \theta = {1 \over 2} \Rightarrow \theta = 60^\circ
Q30
A particle moves in x-y plane according to rule x = asinω\omega t and y = acosω\omega t. The particle follows
A an elliptical path
B a circular path
C a parabolic path
D a straight line path inclined equally to x and y-axes
Correct Answer
Option B
Solution
x=asin⁡ωtx = a\sin \omega t

or

xa=sin⁡ωt{x \over a} = \sin \omega t

....(1)

y=acos⁡ωty = a\cos \omega t

or

ya=cos⁡ωt{y \over a} = \cos \omega t

....(2) Squaring and adding, we get

x2a2+y2a2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{a^2}}} = 1

(∴\therefore

cos⁡2ωt+sin⁡2ωt=1{\cos ^2}\omega t + {\sin ^2}\omega t = 1

) or

x2+y2=a2{x^2} + {y^2} = {a^2}

This is the equation of a circle. Hence particle follows a circular path.

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