Projectile Motion(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

PROJECTILE MOTION MCQs

Total Questions : 45 | Page 1 of 5 pages

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

$\frac{r_{2}}{r_{1}}$
$\sqrt{\frac{r_{2}}{r_{1}}}$
${\frac{(r_{1})}{(r_{2})}}^{2}$
${\frac{(r_{2})}{(r_{1})}}^{2}$

Discuss Question

Answer: Option A. ->

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

:
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 2.

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 3.

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

$u c o s θ$
$u$
$u s i n θ$
$(u c o s θ - u)$

Discuss Question

Answer: Option C. ->

u s i n θ

:
C

Since horizontal component of velocity remain always constant therefore only vertical component of velocity changes.

Initially vertical component $u s i n θ$

Finally it becomes zero. So change in velocity $= u s i n θ$

Question 4.

An aeroplane moving with 150 m/s drops a bomb from a height of 80 m so as to hit a target. What is the distance between the target and the point where the bomb is dropped? (given $g = 10 m / s^{2}$ )

605.3 m
600 m
80 m
230 m

Discuss Question

Answer: Option A. -> 605.3 m
:
A

The horizontal distance covered by bomb,
$B C = v_{H} \times \sqrt{\frac{2 h}{g}} = 150 \sqrt{\frac{2 \times 80}{10}} = 600 m$

$∴$ The distance of target from dropping point
$∴$ The distance of target from dropping point of bomb,
$A C$ = $\sqrt{A B^{2} + B C^{2}}$ = $\sqrt{(80)^{2} + (600)^{2}}$ = $605.3 m$

Question 5.

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

The path will be parabolic

Question 6.

A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of $30^{\circ}$ with the horizontal. The change in momentum (in magnitude) of the body is

24.5 N–s
49.0 N–s
98.0 N–s
50.0 N–s

Discuss Question

Answer: Option B. -> 49.0 N–s
:
B

Change in momentum between complete projectile motion = $2 m u s i n θ = 2 \times 0.5 \times 98 \times s i n 30^{\circ} = 49 N - s$

Question 7.

A particle of mass 100 g is fired with a velocity $20 m s e c^{- 1}$ making an angle of $30^{\circ}$ with the horizontal. When it rises to the highest point of its path then the change in its momentum is

$\sqrt{3}$ $k g m s e c^{- 1}$
$\frac{1}{2} k g m s e c^{- 1}$
$\sqrt{2}$ $k g m s e c^{- 1}$
$1 k g m s e c^{- 1}$

Discuss Question

Answer: Option D. ->

1 k g m s e c^{- 1}

:
D

Horizontal momentum remains always constant. So change in vertical momentum $(Δ \to p)$ = Final vertical momentum – Initial vertical momentum $= - m u s i n θ$

$| Δ P | = 0.1 \times 20 \times s i n 30^{\circ} = 1 k g m / s e c$

Question 8.

Two equal masses (m) are projected at the same angle $(θ)$ from two points separated by their range with equal velocities (v). The total momentum at the point of their collision is

$(Z e r o)$
$2 m v c o s θ$
$- 2 m v c o s θ$
$N o n e o f t h e s e$

Discuss Question

Answer: Option A. ->

(Z e r o)

:
A

Both masses will collide at the highest point of their trajectory with equal and opposite momentum. So net momentum of the system will be zero.

Question 9.

A particle of mass m is projected with velocity v making an angle of $45^{\circ}$ with the horizontal. Themagnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where g = acceleration due to gravity)

$Z e r o$
$\frac{m v^{3}}{(4 \sqrt{2} g)}$
$\frac{m v^{3}}{(\sqrt{2} g)}$
$\frac{m v^{2}}{2 g}$

Discuss Question

Answer: Option B. ->

\frac{m v^{3}}{(4 \sqrt{2} g)}

:
B

$L = \frac{m u^{3} c o s θ s i n^{2} θ}{2 g} = \frac{m v^{3}}{(4 \sqrt{2} g)} [A s θ = 45^{\circ}]$

Question 10.

Four bodies P, Q, R and S are projected with equal velocities having angles of projection $15^{\circ}, 30^{\circ}, 45^{\circ} a n d 60^{\circ}$ with the horizontal respectively. The body having shortest range is