Topic: Motion in A Plane (Test 4)



Topic: Motion in A Plane
Q.1
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 = 
=> u2g= 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

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Q.2
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 , along the radius at every point towards the center
B. 11.9 m , along the radius at every point away from the center
C. 9.9 m , along the radius at every point towards the center
D. 9.9 m , 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, 
Angular frequency, 

Centripetal acceleration, ac = ω2r

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

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Q.3
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


Centripetal acceleration, 


Acceleration due to gravity, g = 9.8 m/s​​​​​​​​​​​2

= 6.4 g


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Q.4
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.

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Q.5
The position of a particle is given by  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  

We know velocity is given by


So 

Acceleration is given by 


So, 


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Q.6

A particle starts from the origin at t = 0 s with a velocity of 10.0  m/s and moves in the x-y plane with a constant acceleration of ( 8.0+ 2.0) m . 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 +axt2

=> 16 = 

t = 2 sec         

y = uyt +   = 10x2 + = 24 m


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Q.7
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, 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° = 
PR = OR tan 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 = 

= 182.24 m/s  m/s


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Q.8

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  to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit ? (Take g = 10 m).


A. at an angle of 20.5 with the vertical, 12 km
B. at an angle of 23.5 with the vertical, 17 km
C. at an angle of 21.5 with the vertical, 15 km
D. at an angle of 19.5 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, = 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 = 
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 = 

= 16006.482 m
≈ 16 km


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Q.9
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 , 58.5with the direction of velocity
B. 0.82 m , 59.5with the direction of velocity
C. 0.86 m , 54.5with the direction of velocity
D. 0.85 m , 56.5with 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 =  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 = 
0.74 = 0.86 ms-2

tan θ = 
where θ is the angle of the resultant with the direction of velocity.
tan θ = = 1.4
θ = tan-1(1.4) = 54.56° with the direction of velocity


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Q.10

A ball is thrown upwards with an initial velocity of 10 m . 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/s​​​​​​2

Let height = h

So we know


=>

=> h= 5.10 m

Also let time taken to reach maximum height = t

Then 

We know 

v = u +at

=> 0 = 10 +(-9.8)t

=> t = 


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