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

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For a given velocity, a projectile has the same range R for two angles of projection. If t₁ and t₂ are the times of flight in the two cases then :

$t_{1} t_{2} \propto R^{2}$

$t_{1} t_{2} \propto R$

$t_{1} t_{2} \propto \frac{1}{R}$

$t_{1} t_{2} \propto \frac{1}{R^{2}}$

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Explanation

For the same range, the angle of projection should be θ and 90–θ.

So, the time of flights $t_{1} = \frac{2 u \sin θ}{g}$ and

$t_{2} = \frac{2 u \sin (90 - θ)}{g} = \frac{2 u \cos θ}{g}$

By multiplying $= t_{1} t_{2} = \frac{4 u^{2} \sin θ \cos θ}{g^{2}}$

$t_{1} t_{2} = \frac{2}{g} \frac{(u^{2} \sin 2 θ)}{g} = \frac{2 R}{g}$ ⇒ $t_{1} t_{2} \propto R$

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A body of mass m is thrown upwards at an angle θ with the horizontal with velocity v. While rising up the velocity of the mass after t seconds will be

$\sqrt{{(v \cos θ)}^{2} + {(v \sin θ)}^{2}}$

$\sqrt{{(v \cos θ - v \sin θ)}^{2} - g t}$

$\sqrt{v^{2} + g^{2} t^{2} - (2 v \sin θ) g t}$

$\sqrt{v^{2} + g^{2} t^{2} - (2 v \cos θ) g t}$

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Explanation

Instantaneous velocity of rising mass after t sec will be $v_{t} = \sqrt{v_{x}^{2} + v_{y}^{2}}$

where $v_{x} = v \cos θ =$ Horizontal component of velocity

$v_{y} = v \sin θ - g t =$ Vertical component of velocity

$v_{t} = \sqrt{{(v \cos θ)}^{2} + {(v \sin θ - g t)}^{2}}$

$v_{t} = \sqrt{v^{2} + g^{2} t^{2} - 2 v \sin θ g t}$

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A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same effort, he throws the ball vertically upwards. The maximum height attained by the ball is

100 m

80 m

60 m

50 m

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Explanation

Maximum range $= \frac{u^{2}}{g} = 100 m$

Maximum height $= \frac{u^{2}}{2 g} = \frac{100}{2} = 50 m$

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A ball is projected with velocity V₀ at an angle of elevation 30°. Mark the correct statement

Kinetic energy will be zero at the highest point of the trajectory

Vertical component of momentum will be conserved

Horizontal component of momentum will be conserved

Gravitational potential energy will be minimum at the highest point of the trajectory

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Explanation

Since horizontal component of velocity is constant, hence momentum is constant.

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Neglecting the air resistance, the time of flight of a projectile is determined by

U_vertical

U_horizontal

U = U²_vertical+ U²_horizontal

U = U(U²_vertical+ U²_horizontal )^1/2

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Explanation

Time of flight = $\frac{2 u \sin θ}{g} = \frac{2 u_{y}}{g} = \frac{2 \times u_{vertical}}{g}$

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A ball is thrown from a point with a speed v₀ at an angle of projection θ. From the same point and at the same instant a person starts running with a constant speed v₀/2 to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?

Yes, 60°

Yes, 30°

Yes, 45°

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Explanation

Person will catch the ball if its velocity will be equal to horizontal component of velocity of the ball.

$\frac{v_{0}}{2} = v_{0} \cos θ$ ⇒ $\cos θ = \frac{1}{2}$ ⇒ $θ = 60 °$

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A stone is thrown at an angle θ to the horizontal reaches a maximum height H. Then the time of flight of stone will be

$\sqrt{\frac{2 H}{g}}$

$2 \sqrt{\frac{2 H}{g}}$

$\frac{2 \sqrt{2 H \sin θ}}{g}$

$\frac{\sqrt{2 H \sin θ}}{g}$

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Explanation

$H = \frac{u^{2} \sin^{2} θ}{2 g}$ and $T = \frac{2 u \sin θ}{g}$ ⇒ $T^{2} = \frac{4 u^{2} \sin^{2} θ}{g^{2}}$

∴ $\frac{T^{2}}{H} = \frac{8}{g}$ ⇒ $T = \sqrt{\frac{8 H}{g}} = 2 \sqrt{\frac{2 H}{g}}$

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The horizontal range of a projectile is $4 \sqrt{3}$ times its maximum height. Its angle of projection will be

45°

60°

90°

30°

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Explanation

$R = 4 H \cot θ,$ if R = $4 \sqrt{3} H$ then $\cot θ = \sqrt{3}$ ⇒ $θ = 30 °$

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The maximum horizontal range of a projectile is 400 m. The maximum value of height attained by it will be

100 m

200 m

400 m

800 m

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Explanation

$R_{\max} = \frac{u^{2}}{g}$ = 400 m $(For θ = 45 °)$

$H_{\max} = \frac{u^{2}}{2 g} = \frac{400}{2} = 200 m$ $(For θ = 90 °)$

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A particle is moving eastwards with velocity of 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is

Zero

$\frac{1}{\sqrt{2}} m / s^{2}$ toward north-west

$\frac{1}{\sqrt{2}} m / s^{2}$ toward north-east

$\frac{1}{2} m / s^{2}$ toward north-west

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