Motion in A Plane - Online Test

# Motion in A Plane # CBSE 11TH PHYSICS Prepare / Learn

Q1. A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

A. 76 m

B. 60 m

C. 65 m

D. 50 m

Answer : Option D
Explaination / Solution:

Maximum horizontal distance, R = 100 m
The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45°, i.e., θ = 45°.
The horizontal range for a projection velocity v, is given by the relation:
R =( u2 Sin 2θ / g)
100 =

\frac{u^{2} S i n 90 °}{g}

=> $\frac{u^{2}}{g}$ = 100 ….(i)
The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity v is zero at the maximum height H.
Acceleration, a = –g
Using the third equation of motion:
v2 – u2 = -2gH

H = \frac{u^{2}}{2 g} = \frac{100}{2} = 50 m

# Motion in A Plane # CBSE 11TH PHYSICS Prepare / Learn

Q2. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

A. 10.2 m

s^{- 2}

, along the radius at every point towards the center

B. 11.9 m

s^{- 2}

, along the radius at every point away from the center

C. 9.9 m

s^{- 2}

, along the radius at every point towards the center

D. 9.9 m

s^{- 2}

, along the radius at every point away from the center

Answer : Option C
Explaination / Solution:

Length of the string, l = 80 cm = 0.8 m
Number of revolutions = 14
Time taken = 25 s
Frequency,

ν = \frac{N u m b e r o f r e v o l u t i o n s}{T i m e t a k e n} = \frac{14}{25} H z

Angular frequency,

ω = 2 π ν

2 \times \frac{22}{7} \times \frac{14}{25} = \frac{88}{25} r a d s^{- 1}

Centripetal acceleration, ac = ω2r
=

(\frac{88}{25})^{2} \times 0.8

= 9.9 ms-2
The direction of centripetal acceleration is always directed along the string, towards the centre, at all points.

# Motion in A Plane # CBSE 11TH PHYSICS Prepare / Learn

Q3. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

A. 10 g

B. 6.4 g

C. 5 g

D. 23 g

Answer : Option B
Explaination / Solution:

Radius of the loop, r = 1 km = 1000 m

Speed of the aircraft, v = 900 km/h

= $900 \times \frac{5}{18} = 250 m / s$

Centripetal acceleration, $a_{c} = \frac{v^{2}}{r}$

= $\frac{(250)^{2}}{1000} = \frac{62500}{1000} = 62.5 m / s^{2}$

Acceleration due to gravity, g = 9.8 m/s2

$\frac{a_{c}}{g} = \frac{62.5}{9.8}$ = 6.4 g

# Motion in A Plane # CBSE 11TH PHYSICS Prepare / Learn

Q4. Which of the following statements not true?

A. The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector

B. The net acceleration of a particle in uniform circular motion is always along the radius of the circle towards the centre

C. The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point

D. The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre

Answer : Option D
Explaination / Solution:

The net acceleration of a particle in circular motion is along the radius of the circle towards the centre Only in case of uniform circular motion.

# Motion in A Plane # CBSE 11TH PHYSICS Prepare / Learn

Q5. The position of a particle is given by

r = 3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k}

Find the velocity and acceleration of the particle.

A. 3.0i^−4.0tj^ , -4.0j^

B. 4.0i^−4.0tj^ , -3.0j^

C. 2.0i^−4.0tj^, -2.0 j

D. 5.0i^−4.0tj^, -3.0 j

Answer : Option A
Explaination / Solution:

Position vector $\vec{r} = 3.0 t \hat{i} - 2.0 t^{2} \hat{j} + 4.0 \hat{k}$

We know velocity is given by

$\vec{v} = \frac{d \vec{r}}{d t}$

So $\vec{v} = 3.0 \hat{i} - 4.0 t \hat{j}$

Acceleration is given by

$\vec{a} = \frac{d \vec{v}}{d t}$

So, $\vec{a} = - 4.0 \hat{j}$

# Motion in A Plane # CBSE 11TH PHYSICS Prepare / Learn

Q6.

A particle starts from the origin at t = 0 s with a velocity of 10.0 $\hat{j}$ m/s and moves in the x-y plane with a constant acceleration of ( 8.0 $\hat{i}$ + 2.0 $\hat{j}$ ) m $s^{- 2}$ . At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?

A. 2.2 s, 22 m

B. 2 s, 24 m

C. 1 s, 14 m

D. 1.5 s, 20 m

Answer : Option B
Explaination / Solution:

u = uxi + uyj = 10j,

ux = 0 and uy = 10 m/s

a = axi + ayj = 8i + 2j,

ax = 8 and ay = 2 m/s2

x = uxt + $\frac{1}{2}$ axt2

=> 16 = $\frac{8 t^{2}}{2}$

t = 2 sec

y = uyt + $\frac{a_{y} t^{2}}{2}$ = 10x2 + $\frac{2 \times 2^{2}}{2}$ = 24 m

# Motion in A Plane # CBSE 11TH PHYSICS Prepare / Learn

Q7. An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s apart is 30

^{\circ}

, what is the speed of the aircraft?

A. 172 m/s

B. 192 m/s

C. 193 m/s

D. 182 m/s

Answer : Option D
Explaination / Solution:

The positions of the observer and the aircraft are shown in the given figure.

Height of the aircraft from ground, OR = 3400 m
Angle subtended between the positions, ∠POQ = 30°
Time = 10 s
In ΔPRO:
tan 15° = $\frac{P R}{O R}$
PR = OR tan 15°
= $3400 \times t a n 15 °$

ΔPRO is similar to ΔRQO.
∴PR = RQ
PQ = PR + RQ
= 2PR = 2 × 3400 tan 15°
= 6800 × 0.268 = 1822.4 m
∴ Speed of the aircraft = $\frac{1822.4}{10}$

= 182.24 m/s $\approx 182$ m/s

# Motion in A Plane # CBSE 11TH PHYSICS Prepare / Learn

Q8.

A fighter plane flying horizontally at an altitude of 1.5 km with speed 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 600 m $s^{- 1}$ to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take g = 10 m $s^{- 2}$ ).

A. at an angle of 20.5

^{\circ}

with the vertical, 12 km

B. at an angle of 23.5

^{\circ}

with the vertical, 17 km

C. at an angle of 21.5

^{\circ}

with the vertical, 15 km

D. at an angle of 19.5

^{\circ}

with the vertical, 16 km

Answer : Option D
Explaination / Solution:

Height of the fighter plane = 1.5 km = 1500 m
Speed of the fighter plane, v = 720 km/h = 200 m/s
Let θ be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.

Muzzle velocity of the gun, u = 600 m/s
Time taken by the shell to hit the plane = t
Horizontal distance travelled by the shell = uxt
Distance travelled by the plane = vt
The shell hits the plane. Hence, these two distances must be equal.
uxt = vt
u Sin θ = v
Sin θ = $\frac{v}{u} = \frac{200}{600}$
θ = Sin-1(0.33) = 19.50

In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell.
∴ H = $\frac{u^{2} s i n^{2} (90 - θ)}{2 g} = \frac{(600)^{2} c o s^{2} θ}{2 g}$
= $\frac{360000 \times c o s^{2} (19.5)}{2 \times 10}$
= 16006.482 m
≈ 16 km

# Motion in A Plane # CBSE 11TH PHYSICS Prepare / Learn

Q9. A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

A. 0.84 m

s^{- 2}

, 58.5

^{\circ}

with the direction of velocity

B. 0.82 m

s^{- 2}

, 59.5

^{\circ}

with the direction of velocity

C. 0.86 m

s^{- 2}

, 54.5

^{\circ}

with the direction of velocity

D. 0.85 m

s^{- 2}

, 56.5

^{\circ}

with the direction of velocity

Answer : Option C
Explaination / Solution:

Speed of the cyclist, v = 27 km/h = 7.5 m/s
Radius of the circular turn, r = 80 m
Centripetal acceleration is given as:
ac = $\frac{v^{2}}{r} = \frac{(7.5)^{2}}{80} =$ 0.7 ms-2

The situation is shown in the given figure:

Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by 0.5 m/s2.

This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.
Since the angle between ac and aT is 900, the resultant acceleration a is given by:
a = $\sqrt{(a_{c})^{2} + (a_{T})^{2}} = \sqrt{(0.7)^{2} + (0.5)^{2}}$
= $\sqrt{0.74}$ = 0.86 ms-2

tan θ = $\frac{a_{c}}{a_{T}}$
where θ is the angle of the resultant with the direction of velocity.
tan θ = $\frac{0.7}{0.5}$ = 1.4
θ = tan-1(1.4) = 54.56° with the direction of velocity

# Motion in A Plane # CBSE 11TH PHYSICS Prepare / Learn

Q10.

A ball is thrown upwards with an initial velocity of 10 m $s^{- 1}$ . Determine the maximum height reached above the thrower’s hand. Determine the time it takes the ball to reach its maximum height.

A. 5.43 m, 0.92 s

B. 5.23 m, 1.12 s

C. 5.10 m, 1.02 s

D. 5.25 m, 0.42 s

Answer : Option C
Explaination / Solution:

Initial velocity u = 10m/s

As at the maximum height ball ll stop so final velocity v = 0 m/s

Only acceleration working on it is acceleration due to gravity g = -9.8m/s2

Let height = h

So we know

$v^{2} - u^{2} = 2 a s$

=> $(0)^{2} - (10)^{2} = 2 \times (- 9.8) h$

=> $h = \frac{- 100}{- 19.6}$ 5.10 m

Also let time taken to reach maximum height = t

Then

We know

v = u +at

=> 0 = 10 +(-9.8)t

=> t = $\frac{- 10}{- 9.8} = 1.02 s$

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