Practice Start Test

Motion in a Plane

JEE Physics · 62 questions · Page 3 of 7 · Click an option or "Show Solution" to reveal answer

Q21
A particle moves such that its position vector r→(t)=cos⁡ωti^+sin⁡ωtj^\overrightarrow r \left( t \right) = \cos \omega t\widehat i + \sin \omega t\widehat j where ω\omega is a constant and t is time. Then which of the following statements is true for the velocity v→(t)\overrightarrow v \left( t \right) and acceleration a→(t)\overrightarrow a \left( t \right) of the particle :
A v→\overrightarrow v and a→\overrightarrow a both are perpendicular to r→\overrightarrow r
B v→\overrightarrow v and a→\overrightarrow a both are parallel to r→\overrightarrow r
C v→\overrightarrow v is perpendicular to r→\overrightarrow r and a→\overrightarrow a is directed towards the origin
D v→\overrightarrow v is perpendicular to r→\overrightarrow r and a→\overrightarrow a is directed away from the origin
Correct Answer
Option C
Solution
r→(t)=cos⁡ωti^+sin⁡ωtj^\overrightarrow r \left( t \right) = \cos \omega t\widehat i + \sin \omega t\widehat j
v→=dr→dt\overrightarrow v = {{d\overrightarrow r } \over {dt}}

=

−ωsin⁡ωt i^+ωcos⁡ωtj^- \omega \sin \omega t\,\widehat i + \omega \cos \omega t\widehat j
a→=dv→dt\overrightarrow a = {{d\overrightarrow v } \over {dt}}

=

−ω2cos⁡ωt i^−ω2sin⁡ωtj^- {\omega ^2}\cos \omega t\,\widehat i - {\omega ^2}\sin \omega t\widehat j

=

−ω2(cos⁡ωt i^+sin⁡ωtj^)- {\omega ^2}\left( {\cos \omega t\,\widehat i + \sin \omega t\widehat j} \right)

=

−ω2r→- {\omega ^2}\overrightarrow r

∴\therefore

a→\overrightarrow a

is antiparallel to

r→\overrightarrow r

and it's direction towards the origin.

v→.r→=\overrightarrow v .\overrightarrow r =
ω(−sin⁡ωtcos⁡ωt+cos⁡ωtsin⁡ωt)\omega \left( { - \sin \omega t\cos \omega t + \cos \omega t\sin \omega t} \right)

= 0 So

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

.

Q22
The initial speed of a projectile fired from ground is u\mathrm{u}. At the highest point during its motion, the speed of projectile is 32u\dfrac{\sqrt{3}}{2} u. The time of flight of the projectile is :
A ug\dfrac{u}{g}
B 2ug\dfrac{2u}{g}
C u2g\dfrac{u}{2g}
D 3ug\dfrac{\sqrt3u}{g}
Correct Answer
Option A
Solution

ucos⁡θ=32uu \cos \theta=\dfrac{\sqrt{3}}{2} u ⇒\Rightarrow cos⁡θ=32\cos \theta=\dfrac{\sqrt{3}}{2} ⇒\Rightarrow θ=30∘\theta=30^{\circ} Time of flight =2usin⁡θg=(ug)=\dfrac{2 u \sin \theta}{g}=\left(\dfrac{u}{g}\right)

Q23
The trajectory of a projectile in a vertical plane is y = α\alphax −- β\betax2, where α\alpha and β\beta are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection θ\theta and the maximum height attained H are respectively given by :
A tan⁡−1α,α24β{\tan ^{ - 1}}\alpha ,{{{\alpha ^2}} \over {4\beta }}
B tan⁡−1α,4α2β{\tan ^{ - 1}}\alpha ,{{4{\alpha ^2}} \over \beta }
C tan⁡−1(βα),α2β{\tan ^{ - 1}}\left( {{\beta \over \alpha }} \right),{{{\alpha ^2}} \over \beta }
D tan⁡−1β,α22β{\tan ^{ - 1}}\beta ,{{{\alpha ^2}} \over {2\beta }}
Correct Answer
Option A
Solution

y = α\alphax −- β\betax2 comparing with trajectory equation

y=xtan⁡θ−12gx2u2cos⁡2θy = x\tan \theta - {1 \over 2}{{g{x^2}} \over {{u^2}{{\cos }^2}\theta }}
tan⁡θ=α⇒θ=tan⁡−1α\tan \theta = \alpha \Rightarrow \theta = {\tan ^{ - 1}}\alpha
β=12gu2cos⁡2θ\beta = {1 \over 2}{g \over {{u^2}{{\cos }^2}\theta }}

⇒\Rightarrow

u2=g2βcos⁡2θ{u^2} = {g \over {2\beta {{\cos }^2}\theta }}

Maximum height H :

H=u2sin⁡2θ2g=g2βcos⁡2θsin⁡2θ2gH = {{{u^2}{{\sin }^2}\theta } \over {2g}} = {g \over {2\beta {{\cos }^2}\theta }}{{{{\sin }^2}\theta } \over {2g}}

⇒\Rightarrow

H=tan⁡2θ4β=α24βH = {{{{\tan }^2}\theta } \over {4\beta }} = {{{\alpha ^2}} \over {4\beta }}
Q24
A mosquito is moving with a velocity v→=0.5t2i^+3tj^+9k^\overrightarrow v = 0.5{t^2}\widehat i + 3t\widehat j + 9\widehat k m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2 s?
A tan⁡−1(856){\tan ^{ - 1}}\left( {{\sqrt {85} } \over 6}\right) from y-axis
B tan⁡−1(52){\tan ^{ - 1}}\left( {{5 \over 2}} \right) from y-axis
C tan⁡−1(23){\tan ^{ - 1}}\left( {{2 \over 3}} \right) from x-axis
D tan⁡−1(52){\tan ^{ - 1}}\left( {{5 \over 2}} \right) from x-axis
Correct Answer
Option A
Solution
v→=(0.5t2i^+3tj^+9k^)\overrightarrow v = (0.5{t^2}\widehat i + 3t\widehat j + 9\widehat k)

m/s At t = 2 s

v→=(2i^+6j^+9k^)\overrightarrow v = (2\widehat i + 6\widehat j + 9\widehat k)

Direction cosine along y-axis,

cosθ=(v.j^)92+62+22=6121=611cos\theta = {{(v.\widehat j)} \over {\sqrt {{9^2} + {6^2} + {2^2}} }} = {6 \over {\sqrt {121} }} = {6 \over {11}}

∴\therefore

sin⁡θ=8511\sin \theta = {{\sqrt {85} } \over {11}}

and

tan⁡θ=856\tan \theta = {{\sqrt {85} } \over 6}

∴\therefore Mosquito make angle

tan⁡−1(856){\tan ^{ - 1}}\left( {{\sqrt {85} } \over 6}\right)

from y-axis.

Q25
A butterfly is flying with a velocity 424\sqrt 2 m/s in North-East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 seconds is :
A 12212\sqrt 2 m
B 20 m
C 3 m
D 15 m
Correct Answer
Option D
Solution

The given situation can be represented as In the above figure, v1 is the speed of wind and v21 is the speed of butterfly with respect to wind.

So, v21 can be given as

v21=42cos⁡45∘i^+42sin⁡45∘j^{v_{21}} = 4\sqrt 2 \cos 45^\circ \widehat i + 4\sqrt 2 \sin 45^\circ \widehat j
=42×12i^+42×12j^=4i^+4j^= 4\sqrt 2 \times {1 \over {\sqrt 2 }}\widehat i + 4\sqrt 2 \times {1 \over {\sqrt 2 }}\widehat j = 4\widehat i + 4\widehat j

and v1 can be given as

v1=−j^{v_1} = - \widehat j

∴\therefore Velocity of butterfly can be given as

v2=v1+v21=4i^+4j^−j^=4i^+3j^{v_2} = {v_1} + {v_{21}} = 4\widehat i + 4\widehat j - \widehat j = 4\widehat i + 3\widehat j

∴\therefore Displacement of butterfly,

D=v2×tD = {v_2} \times t
=(4i^+3j^)×3=12i^+9j^= (4\widehat i + 3\widehat j) \times 3 = 12\widehat i + 9\widehat j

∴\therefore Magnitude of displacement,

∣D∣=122+92=15\left| D \right| = \sqrt {{{12}^2} + {9^2}} = 15

m

Q26
A bomb is dropped by fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a :
A hyperbola
B parabola in the direction of motion of plane
C straight line vertically down the plane
D parabola in a direction opposite to the motion of plane
Correct Answer
Option C
Solution

The correct answer is Option C, a straight line vertically down the plane.

Here's why: Frame of Reference: The key to understanding projectile motion is considering the frame of reference.

The observer in the plane is in a moving frame of reference.

From their perspective, they are at rest, and the bomb has the same horizontal velocity as the plane.

Gravity's Influence: The only force acting on the bomb after it's released is gravity.

Gravity acts vertically downwards.

Resultant Motion: Because the bomb has the same horizontal velocity as the plane and is only affected by gravity vertically, it appears to the observer in the plane to fall straight down.

Let's eliminate the other options: Option A (Hyperbola): A hyperbolic trajectory occurs when an object is influenced by two forces acting in different directions, resulting in a path that gets progressively straighter.

This doesn't apply to the bomb scenario.

Option B and D (Parabola): A parabolic trajectory is observed from a stationary frame of reference on the ground.

In this case, the horizontal motion of the bomb combined with the vertical acceleration due to gravity creates a parabolic path.

Q27
A player kicks a football with an initial speed of 25 ms−-1 at an angle of 45∘^\circ from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion ? (Take g = 10 ms−-2)
A hmax = 10 m T = 2.5 s
B hmax = 15.625 m T = 3.54 s
C hmax = 15.625 m T = 1.77 s
D hmax = 3.54 m T = 0.125 s
Correct Answer
Option C
Solution
H=U2sin⁡2θ2gH = {{{U^2}{{\sin }^2}\theta } \over {2g}}
=(25)2.(sin⁡45)22×10= {{{{(25)}^2}.{{(\sin 45)}^2}} \over {2 \times 10}}

= 15.625 m

T=Usin⁡θgT = {{U\sin \theta } \over g}
=25×sin⁡45∘10= {{25 \times \sin 45^\circ } \over {10}}

= 2.5 ×\times 0.7 = 1.77 s

Q28
A helicopter is flying horizontally with a speed 'v' at an altitude 'h' has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped?
A 2ghv2+1h2\sqrt {{{2gh{v^2} + 1} \over {{h^2}}}}
B 2ghv2+h2\sqrt {2gh{v^2} + {h^2}}
C 2v2hg+h2\sqrt {{{2{v^2}h} \over g} + {h^2}}
D 2ghv2+h2\sqrt {{{2gh} \over {{v^2}}} + {h^2}}
Correct Answer
Option C
Solution
R=2hg. vR = \sqrt {{{2h} \over g}} .\,v
D=R2+h2D = \sqrt {{R^2} + {h^2}}
=(2hg. v)2+h2= \sqrt {{{\left( {\sqrt {{{2h} \over g}} .\,v} \right)}^2} + {h^2}}
D=2hv2g+h2D = \sqrt {{{2h{v^2}} \over g} + {h^2}}

Option (c) is correct.

Q29
The ranges and heights for two projectiles projected with the same initial velocity at angles 42∘^\circ and 48∘^\circ with the horizontal are R1, R2 and H1, H2 respectively. Choose the correct option :
A R1 > R2 and H1 = H2
B R1 = R2 and H1 < H2
C R1 < R2 and H1 < H2
D R1 = R2 and H1 = H2
Correct Answer
Option B
Solution

Here, two projectiles are projected at angles 42

∘^\circ

and 48

∘^\circ

with same initial velocity. As we know the expression of range of projectile, Range

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

AT θ\theta1 = 42

∘^\circ

, Range,

R1=u2sin⁡2(42)∘g=0.99u2g{R_1} = {{{u^2}\sin 2(42)^\circ } \over g} = {{0.99{u^2}} \over g}

At θ\theta2 = 48

∘^\circ

Range,

R2=u2sin⁡2(48)∘g=0.99u2g{R_2} = {{{u^2}\sin 2(48)^\circ } \over g} = {{0.99{u^2}} \over g}

The range of the projectile is same for the two projectiles.

Therefore, R1 = R2 Now, as we know the expression of height of the projectile,

Hmax⁡=u2sin⁡θ2g{H_{\max }} = {{{u^2}\sin \theta } \over {2g}}

At 42

∘^\circ

,

Hmax⁡=H1=u2sin⁡42∘2g=0.669u22g{H_{\max }} = {H_1} = {{{u^2}\sin 42^\circ } \over {2g}} = {{0.669{u^2}} \over {2g}}

At 48

∘^\circ

,

Hmax⁡=H2=u2sin⁡48∘2g=0.743u22g{H_{\max }} = {H_2} = {{{u^2}\sin 48^\circ } \over {2g}} = {{0.743{u^2}} \over {2g}}

Higher the value of θ\theta higher the value of maximum height. Therefore, H1 2.

Q30
A person can throw a ball upto a maximum range of 100 m. How high above the ground he can throw the same ball?
A 25 m
B 50 m
C 100 m
D 200 m
Correct Answer
Option B
Solution

To determine how high a person can throw a ball given the maximum range, we need to use the principles of projectile motion in physics.

The maximum range of a projectile is given by the formula:

R=v02sin⁡(2θ)gR = \frac{{v_0^2 \sin(2\theta)}}{g}

where:

RR

is the range

v0v_0

is the initial velocity θ\theta is the angle of projection

gg

is the acceleration due to gravity (approximately

9.8 m/s29.8 \, \text{m/s}^2

) The maximum range is achieved when

θ=45∘\theta = 45^\circ

, thus

sin⁡(2θ)=sin⁡(90∘)=1\sin(2\theta) = \sin(90^\circ) = 1

. Given that the maximum range

R=100 mR = 100 \, \text{m}

, we can rewrite the range formula as:

100=v02⋅19.8100 = \frac{{v_0^2 \cdot 1}}{9.8}

Solving for

v02v_0^2

:

v02=100×9.8=980v_0^2 = 100 \times 9.8 = 980

Next, the maximum height

HH

reached by the ball can be calculated using the vertical component of the velocity.

The formula for the maximum height is:

H=v02sin⁡2(θ)2gH = \frac{{v_0^2 \sin^2(\theta)}}{2g}

Here, for a vertical throw,

θ=90∘\theta = 90^\circ

, and thus

sin⁡(90∘)=1\sin(90^\circ) = 1

. Substituting the values, we get:

H=v02⋅12⋅9.8H = \frac{{v_0^2 \cdot 1}}{2 \cdot 9.8}

Using

v02=980v_0^2 = 980

, we have:

H=9802×9.8=98019.6=50 mH = \frac{980}{2 \times 9.8} = \frac{980}{19.6} = 50 \, \text{m}

Therefore, the correct answer is: Option B: 50 m

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