Motion in a Plane
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A particle moves along a circle of radius 
20 
m with constant tangential acceleration. π
If the velocity of the particle is 80 m/s at the end of the second revolution after motion has begun, the tangential acceleration is
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Circumference = 2πr = 2π × 20 = 40 m π
Distance travelled in 2 revolutions = 2 × 40 = 80 m
Initial velocity = u = 0 Final velocity v = 80m/sec
Applying the formula, v2 = u2 + 2as
(80)2 = 02 + 2 × a × 80 ⇒ a = 40 m/sec2Correct Option: B
Circumference = 2πr = 2π × 20 = 40 m π
Distance travelled in 2 revolutions = 2 × 40 = 80 m
Initial velocity = u = 0 Final velocity v = 80m/sec
Applying the formula, v2 = u2 + 2as
(80)2 = 02 + 2 × a × 80 ⇒ a = 40 m/sec2
- A stone is tied to a string of length l and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in velocity as it reaches a position where the string is horizontal (g being acceleration due to gravity) is
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Wmg = ∆K
⇒ – mgl = ½ mv2 – ½ mu2
or, mv2 = m( u2 – 2gl )
or , v→ = √u² - 2gl ĵ
u→ = uî
∴ v→ - u→ = √u² - 2gl ĵ - uî
∴ | v→ - u→ | = [ (u² - 2gl) + u² ]1 / 2 = √2(u² - gl)Correct Option: B
Wmg = ∆K
⇒ – mgl = ½ mv2 – ½ mu2
or, mv2 = m( u2 – 2gl )
or , v→ = √u² - 2gl ĵ
u→ = uî
∴ v→ - u→ = √u² - 2gl ĵ - uî
∴ | v→ - u→ | = [ (u² - 2gl) + u² ]1 / 2 = √2(u² - gl)
- The circular motion of a particle with constant speed is
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In circular motion of a particle with constant speed, particle repeats its motion after a regular interval of time but does not oscillate about a fixed point. So, motion of particle is periodic but not simple harmonic.
Correct Option: A
In circular motion of a particle with constant speed, particle repeats its motion after a regular interval of time but does not oscillate about a fixed point. So, motion of particle is periodic but not simple harmonic.
- A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone?
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ar = ω2R & at = dv = 0 dt ar = (2πn)2R = 4π2n2R2 = 4π2 22 2 (1)2 44 
anet = ar = π2ms–2 and direction along the radius towards the centre.
Correct Option: A
ar = ω2R & at = dv = 0 dt ar = (2πn)2R = 4π2n2R2 = 4π2 22 2 (1)2 44
anet = ar = π2ms–2 and direction along the radius towards the centre.
- Two boys are standing at the ends A and B of a ground where AB = a. The boy at B starts running in a direction perpendicular to AB with velocity v1. The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is
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Velocity of A relative to B is given by
V→A / B = v→A - v→B = v→ - v→1 ......(1)
By taking x-components of equation (1), we get0 = v sin θ - v1 ⇒ sin θ = v1 .........(2) v
By taking Y-components of equation (1), we get
vy = v cos θ ........(3)
Time taken by boy at A to catch the boy at B is given byt = Relative displacement along Y - axis Relative velocity along Y - axis = a = a v cos θ v . √1 - sin² θ 
[From equation (1)]
Correct Option: D
Velocity of A relative to B is given by
V→A / B = v→A - v→B = v→ - v→1 ......(1)
By taking x-components of equation (1), we get0 = v sin θ - v1 ⇒ sin θ = v1 .........(2) v
By taking Y-components of equation (1), we get
vy = v cos θ ........(3)
Time taken by boy at A to catch the boy at B is given byt = Relative displacement along Y - axis Relative velocity along Y - axis = a = a v cos θ v . √1 - sin² θ
[From equation (1)]
