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Motion in a Lane MCQs for NEET — Physics Questions with Answers

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A projectile fired with initial velocity u at some angle θ has a range R. If the initial velocity be doubled at the same angle of projection, then the range will be

2 R

R/2

4 R

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Explanation

$R = \frac{u^{2} \sin 2 θ}{g} ∴ R \propto u^{2}$ . If initial velocity be doubled then range will become four times.

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If the initial velocity of a projectile be doubled, keeping the angle of projection same, the maximum height reached by it will

Remain the same

Be doubled

Be quadrupled

Be halved

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Explanation

$H = \frac{u^{2} \sin^{2} θ}{2 g} ∴ H \propto u^{2} .$

If initial velocity be doubled then maximum height reached by the projectile will quadrupled.

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In the motion of a projectile freely under gravity, its

Total energy is conserved

Momentum is conserved

Energy and momentum both are conserved

None is conserved

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Explanation

An external force by gravity is present throughout the motion so momentum will not be conserved.

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The range of a projectile for a given initial velocity is maximum when the angle of projection is 45°. The range will be minimum, if the angle of projection is

90°

180°

60°

75°

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Explanation

Range $= \frac{u^{2} \sin 2 θ}{g}$ ; when $θ = 90 °$ , R = 0 i.e. the body will fall at the point of projection after completing one dimensional motion under gravity.

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A ball is thrown upwards and it returns to ground describing a parabolic path. Which of the following remains constant

Kinetic energy of the ball

Speed of the ball

Horizontal component of velocity

Vertical component of velocity

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Explanation

Because there is no accelerating or retarding force available in horizontal motion.

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At the top of the trajectory of a projectile, the directions of its velocity and acceleration are

Perpendicular to each other

Parallel to each other

Inclined to each other at an angle of 45°

Antiparallel to each other

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Explanation

Direction of velocity is always tangent to the path so at the top of trajectory, it is in horizontal direction and acceleration due to gravity is always in vertically downward direction. It means angle between $\vec{v}$ and $\vec{g}$ are perpendicular to each other.

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An object is thrown along a direction inclined at an angle of 45° with the horizontal direction. The horizontal range of the particle is equal to

Vertical height

Twice the vertical height

Thrice the vertical height

Four times the vertical height

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Explanation

$R = 4 H \cot θ$ if $θ = 45 °$ then $R = 4 H \cot (45 °) = 4 H$

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The height y and the distance x along the vertical plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = (8 t - 5 t^{2})$ meter and x = 6t meter, where t is in second. The velocity with which the projectile is projected is

8 m/sec

6 m/sec

10 m/sec

Not obtainable from the data

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Explanation

$v_{y} = \frac{d y}{d t} = 8 - 10 t$ , $v_{x} = \frac{d x}{d t} = 6$

at the time of projection i.e. $v_{y} = \frac{d y}{d t} = 8$ and v_x = 6

$∴ v = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{6^{2} + 8^{2}} = 10 m / s$

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The range of a particle when launched at an angle of 15° with the horizontal is 1.5 km. What is the range of the projectile when launched at an angle of 45° to the horizontal

1.5 km

3.0 km

6.0 km

0.75 km

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Explanation

$R_{15 °} = \frac{u^{2} \sin (2 \times 15 °)}{g} = \frac{u^{2}}{2 g} = 1 .5 k m$

$R_{45 °} = \frac{u^{2} \sin (2 \times 45 °)}{g} = \frac{u^{2}}{g} = 1 .5 \times 2 = 3 k m$

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Galileo writes that for angles of projection of a projectile at angles $(45 + θ)$ and $(45 - θ)$ , the horizontal ranges described by the projectile are in the ratio of (if $θ \leq 45$ )

2 : 1

1 : 2

1 : 1

2 : 3

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Explanation

For angle $(45 ° - θ)$ , $R = \frac{u^{2} \sin (90 ° - 2 θ)}{g} = \frac{u^{2} \cos 2 θ}{g}$

For angle $(45 ° + θ)$ , $R = \frac{u^{2} \sin (90 ° + 2 θ)}{g} = \frac{u^{2} \cos 2 θ}{g}$

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