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

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A projectile thrown with a speed v at an angle θ has a range R on the surface of earth. For same v and θ, its range on the surface of moon will be (acceleration due to gravity on moon= $\frac{g}{6}$ ):

R/6

6 R

R/36

36 R

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Explanation

Range is given by $R = \frac{u^{2} \sin 2 θ}{g}$

On moon $g_{m} = \frac{g}{6}$ . Hence R_m = 6 R

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A ball is projected with kinetic energy E at an angle of 45° to the horizontal. At the highest point during its flight, its kinetic energy will be

Zero

$\frac{E}{2}$

$\frac{E}{\sqrt{2}}$

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Explanation

$E = \frac{1}{2} {mu}^{2} At highest point v = u \cos 45 ° E' = \frac{1}{2} {mv}^{2} = \frac{1}{2} m {(u \cos 45 °)}^{2} = \frac{1}{2} {mu}^{2} \times \frac{1}{2} = \frac{E}{2}$

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At the top of the trajectory of a projectile, the magnitude of the acceleration is

Maximum

Minimum

Zero

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Explanation

Acceleration through out the projectile motion remains constant and equal to g.

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A body is projected at such an angle that the horizontal range is three times the greatest height. The angle of projection is

$25^{o} 8^{'}$

$33^{o} 7^{'}$

$42^{o} 8^{'}$

$53^{o} 8^{'}$

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Explanation

$R = 4 H \cot θ,$ if R = 3H then $\cot θ = \frac{3}{4}$ ⇒ $θ = 53 ° 8'$

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Two bodies are projected with the same velocity. If one is projected at an angle of 30° and the other at an angle of 60° to the horizontal, the ratio of the maximum heights reached is

3 : 1

1 : 3

1 : 2

2 : 1

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Explanation

As $H = \frac{u^{2} \sin^{2} θ}{2 g}$ ∴ $\frac{H_{1}}{H_{2}} = \frac{\sin^{2} θ_{1}}{\sin θ_{2}} = \frac{\sin^{2} 30 °}{\sin^{2} 60}$ = $\frac{1 / 4}{3 / 4} = \frac{1}{3}$

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If a body A of mass M is thrown with velocity V at an angle of 30° to the horizontal and another body B of the same mass is thrown with the same speed at an angle of 60° to the horizontal. The ratio of horizontal range of A to B will be

1 : 3

1 : 1

$1 : \sqrt{3}$

$\sqrt{3} : 1$

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Explanation

For complementary angles range will be equal.

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Four bodies P, Q, R and S are projected with equal velocities having angles of projection 15^o, 30^o, 45^o and 60^o with the horizontal respectively. The body having shortest range is

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Explanation

When the angle of projection is very far from 45° then range will be minimum.

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Which of the following sets of factors will affect the horizontal distance covered by an athlete in a long–jump event

Speed before he jumps and his weight

The direction in which he leaps and the initial speed

The force with which he pushes the ground and his speed

None of these

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Explanation

Range $= \frac{u^{2} \sin 2 θ}{g} .$

It is clear that range is proportional to the direction (angle) and the initial speed.

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In a projectile motion, velocity at maximum height is

$\frac{u \cos θ}{2}$

$u \cos θ$

$\frac{u \sin θ}{2}$

None of these

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Explanation

Only horizontal component of velocity $(u \cos θ)$ .

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The equation of motion of a projectile are given by x = 36 t metre and 2y = 96 t – 9.8 t² metre. The angle of projection is

$\sin^{- 1} (\frac{4}{5})$

$\sin^{- 1} (\frac{3}{5})$

$\sin^{- 1} (\frac{4}{3})$

$\sin^{- 1} (\frac{3}{4})$

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Explanation

x = 36 t ∴ $v_{x} = \frac{d x}{d t} = 36 m / s$

$y = 48 t - 4 .9 t^{2}$ ∴ $v_{y} = 48 - 9 .8 t$

at t = 0, v_x = 36 and v_y = 48 m/s

So, angle of projection $θ = \tan^{- 1} (\frac{v_{y}}{v_{x}}) = \tan^{- 1} (\frac{4}{3})$

Or $θ = \sin^{- 1} (4 / 5)$

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