Projectile Motion(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

PROJECTILE MOTION MCQs

Total Questions : 45 | Page 3 of 5 pages

Question 21. From the following displacement-time graph find out the velocity of a moving body

1√3m/s
3m/s
√3 m/s
13

Discuss Question

Answer: Option C. -> √3 m/s
:
C
In first instant you will apply u=tan

θ

and say v=tan

30^{\circ} = \frac{1}{\sqrt{3}} m / s

But it is wrong because the formula

v = t a n θ

is valid when angle is measured
from time axis.
Here angle is taken from displacement axis. So angle from time axis

90^{\circ} - 30^{\circ} = 60^{\circ}

Now v=tan

60^{\circ} = \sqrt{3}

Question 22. A projectile is thrown into space so as to have maximum horizontal range R. Taking the point of projection as origin, the co-ordinates of the point where the speed of the particle is minimum are

(R,R)
(R,R2)
(R2,R4)
(R,R4)

Discuss Question

Answer: Option C. -> (R2,R4)
:
C
For maximum horizontal Range

θ = 45^{\circ}

From R = 4H

c o t θ

= 4H [As

θ = 45^{\circ}

, for maximum range.]
Speed of the particle will be minimum at the highest point of parabola. So the co-ordinate of the highest point will be

(\frac{R}{2}, \frac{R}{4})

A Projectile Is Thrown Into Space So As To Have Maximum Hori...

Question 23. A car moving on a horizontal road may be thrown out of the road in taking a turn

By the gravitational force
Due to lack of sufficient centripetal force
Due to rolling frictional force between tyre and road
Due to the reaction of the ground

Discuss Question

Answer: Option B. -> Due to lack of sufficient centripetal force
:
B
Due to lack of sufficient friction which provides the centripetal force.

Question 24. Two particles of equal masses are revolving in circular paths of radii

r_{1}

and

r_{2}

respectively with the same speed. The ratio of their centripetal forces is

r2r1
√r2r1
(r1)(r2)2
(r2)(r1)2

Discuss Question

Answer: Option A. -> r2r1
:
A
F =

\frac{m v^{2}}{r}

.If m and v are constants then F

\propto

\frac{1}{r}

\frac{F_{1}}{F_{2}}

\frac{r_{2}}{r_{1}}

Question 25. A particle P is projected with velocity

u_{1}

at an angle of

30^{\circ}

with the horizontal. Another particle Q is thrown vertically upwards with velocity

u_{2}

from a point vertically below the highest point of path of P. The necessary condition for the two particles to collide at the highest point is

A Particle P Is Projected With Velocity U1 At An Angle Of 30...

u1=u2
u1=2u2
u1=u22
u1=4u2

Discuss Question

Answer: Option B. -> u1=2u2
:
B
Both particle collide at the highest point it means the vertical distance travelled by both the particle will be equal, i.e. the vertical component of velocity of both particle will be equal

u_{1} s i n 30^{\circ} = u_{2} \Rightarrow \frac{u_{1}}{2} = u_{2} ∴ u_{1} = 2 u_{2}

Question 26. A particle is thrown with velocity u at an angle

α

from the horizontal. Another particle is thrown with the same velocity at an angle

α

from the vertical. The ratio of times of flight of two particles will be

tan 2 α:1
cot 2 α:1
tan α:1
cot α:1

Discuss Question

Answer: Option C. -> tan α:1
:
C
For first particles angle of projection from the horizontal is a. So

T_{1} = \frac{2 u s i n α}{g}

For second particle angle of projection from the vertical is

α

. it mean from the horizontal is

90 - α

∴ T_{2} = \frac{2 u s i n (90 - α)}{g} = \frac{2 u c o s α}{g} .

So ratio of time of flight

\frac{T_{1}}{T_{2}} = t a n α

Question 27. The friction of the air causes vertical retardation equal to one tenth of the acceleration due to gravity (Take g =

10 m s^{- 2}

). The time of flight will be decreased by

Discuss Question

Answer: Option C. -> 9%
:
C

T = \frac{2 u s i n θ}{g} ∴ \frac{T_{1}}{T_{2}} = \frac{g_{2}}{g_{1}} = \frac{g + \frac{g}{10}}{g} = \frac{11}{10}

Fractional decrease in time of flight

= \frac{T_{1} - T_{2}}{T_{1}} = \frac{1}{11}

Percentage decrease

= 9 %

Question 28. A particle moves in a plane with constant acceleration in a direction different from the initial velocity. The path of the particle will be

A straight line
An arc of a circle
A parabola
An ellipse

Discuss Question

Answer: Option C. -> A parabola
:
C
A parabola, because the component of velocity perpendicular to acceleration will remain same.

Question 29. A projectile is projected from the base of an inclined plane of inclination '

α

'. The angle of projection of the projectile with the ground is

θ

. If it strikes the incline plane at it's maximum height from the ground, then, (assume the plane of motion contains the line of greatest slope on the incline plane)

tan θ = tanα2
tan α = tanθ2
tan α = tanθ3
We can not define a relationship between θ and α as it will depend one the speed of projection

Discuss Question

Answer: Option B. -> tan α = tanθ2
:
B
From triangle,

A Projectile Is Projected From The Base Of An Inclined Plane...

\frac{b}{x}

= tan

α

tan

α

\frac{h_{m a x}}{\frac{R}{2}}

\frac{t a n (θ)}{2}

Question 30. A projectile is fired with velocity

u

making angle

θ

with the horizontal. What is the change in velocity when it is at the highest point

u cos θ
u
u sin θ
(u cos θ−u)

Discuss Question

Answer: Option C. -> u sin θ
:
C
Since horizontal component of velocity remain always constant therefore only vertical component of velocity changes.
Initially vertical component