Motion in a Plane
- A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be :
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The magnitude of the resultant velocity at the point of projection and the landing point is same.
Clearly, change in momentum along horizontal (i.e along x-axis) = mv cos θ – mv cos θ = 0
Change in momentum along vertical (i.e. along y–axis) = mv sinθ – (–mv sinθ) = 2 mvsinθ = 2mv × sin 45°= 2mv × 1 = √2 mv √2
Hence, resultant change in momentum = √2 mv
Correct Option: C
The magnitude of the resultant velocity at the point of projection and the landing point is same.
Clearly, change in momentum along horizontal (i.e along x-axis) = mv cos θ – mv cos θ = 0
Change in momentum along vertical (i.e. along y–axis) = mv sinθ – (–mv sinθ) = 2 mvsinθ = 2mv × sin 45°= 2mv × 1 = √2 mv √2
Hence, resultant change in momentum = √2 mv
- A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is
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H = u2 sin2 45° = u2 .....(1) 2g 4g R = u2 sin 90° = u2 g g R = u2 .....(2) 2 2g ∴ tan α = H R / 2 = u2 4g = 1 u2 2 2g ∴ α = tan-1 
1 
2 
Correct Option: B
H = u2 sin2 45° = u2 .....(1) 2g 4g R = u2 sin 90° = u2 g g R = u2 .....(2) 2 2g ∴ tan α = H R / 2 = u2 4g = 1 u2 2 2g ∴ α = tan-1 1 2
- A missile is fired for maximum range with an initial velocity of 20 m/s. If g = 10 m/s2 , the range of the missile is
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For maximum range, the angle of projection, θ = 45°
∴ R = u2 sin 2θ g = (20)2 sin (2 × 45°) 10 = 400 × 1 = 40 m 10 Correct Option: A
For maximum range, the angle of projection, θ = 45°
∴ R = u2 sin 2θ g = (20)2 sin (2 × 45°) 10 = 400 × 1 = 40 m 10
- The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectiles is :
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Horizontal range = u2 sin 2θ .....(1) g Maximum height = u2 sin2 θ .....(2) 2g
According to the problem
R = Hu2 sin 2θ = u2 sin2 θ g 2g ⇒ 2 sin θ cos θ = sin2 θ 2 2 cos θ = sin θ 2 ⇒ cot θ = 1 4
⇒ tan θ = 4
⇒ θ = [ tan-1 (4) ]
Correct Option: B
Horizontal range = u2 sin 2θ .....(1) g Maximum height = u2 sin2 θ .....(2) 2g
According to the problem
R = Hu2 sin 2θ = u2 sin2 θ g 2g ⇒ 2 sin θ cos θ = sin2 θ 2 2 cos θ = sin θ 2 ⇒ cot θ = 1 4
⇒ tan θ = 4
⇒ θ = [ tan-1 (4) ]
- The velocity of a projectile at the initial point A is ( 2î - 3ĵ )m/s. It’s velocity (in m/s) at point B is
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At point B the direction of velocity component of the projectile along Y - axis reverses.
Hence, V→B = 2î - 3ĵCorrect Option: B
At point B the direction of velocity component of the projectile along Y - axis reverses.
Hence, V→B = 2î - 3ĵ
