Two point masses 1 and 2 move with uniform velocities v → 1 a n d v → 2 respectively. Their initial position vectors are r → 1 a n d r → 2 respectively. Which of the following should be satisfied for the collision of the point masses ?

r 2 → - r 1 → r 2 → - r 1 → = v 2 → - v 1 → v 2 → - v 1 →

r 2 → - r 1 → r 2 → + r 1 → = v 2 → + v 1 → v 2 → - v 1 →

r 2 → - r 1 → r 2 → + r 1 → = v 2 → - v 1 → v 2 → + v 1 →

r 2 → + r 1 → r 2 → + r 1 → = v 2 → - v 1 → v 2 → + v 1 →

Physics - Motion Plane - Quiz-6

A particle starts from rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be

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$110 {ms}^{- 1}$
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$55 {ms}^{- 1}$
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$550 {ms}^{- 1}$
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$660 {ms}^{- 1}$

The given graph shows the variation of velocity with displacement. Which one of the graph given below correctly represents the variation of acceleration with displacement?

A particle moves along a parabolic path y = 9x² in such a way that the x component of the velocity remains constant and has a value of $\frac{1}{3} {ms}^{- 1}$ . It can be deduced that the acceleration of the particle will be

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$\frac{1}{3} \hat{j} {ms}^{- 2}$
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$3 \hat{j} {ms}^{- 2}$
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$\frac{2}{3} \hat{j} {ms}^{- 2}$
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$2 \hat{j} {ms}^{- 2}$

The relation between time and distance is $t = {αx}^{2} + βx$ , where $α$ and $β$ are constants. The retardation is

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$2 {αv}^{3}$
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$2 {βv}^{3}$
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$2 {αβv}^{3}$
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$2 β^{2} v^{3}$

A point moves in a straight line under the retardation ${av}^{2}$ . If the initial velocity is u, the distance covered in 't' second is-

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aut
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$\frac{1}{a} \ln (aut)$
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$\frac{1}{a} \ln (1 + aut)$
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$a \ln (aut)$

Two point masses 1 and 2 move with uniform velocities ${\vec{v}}_{1} a n d {\vec{v}}_{2}$ respectively. Their initial position vectors are ${\vec{r}}_{1} a n d {\vec{r}}_{2}$ respectively. Which of the following should be satisfied for the collision of the point masses ?

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$\frac{\vec{r_{2}} - \vec{r_{1}}}{|\vec{r_{2}} + \vec{r_{1}}|} = \frac{\vec{v_{2}} + \vec{v_{1}}}{|\vec{v_{2}} - \vec{v_{1}}|}$
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$\frac{\vec{r_{2}} - \vec{r_{1}}}{|\vec{r_{2}} - \vec{r_{1}}|} = \frac{\vec{v_{2}} - \vec{v_{1}}}{|\vec{v_{2}} - \vec{v_{1}}|}$
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$\frac{\vec{r_{2}} - \vec{r_{1}}}{|\vec{r_{2}} + \vec{r_{1}}|} = \frac{\vec{v_{2}} - \vec{v_{1}}}{|\vec{v_{2}} + \vec{v_{1}}|}$
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$\frac{\vec{r_{2}} + \vec{r_{1}}}{|\vec{r_{2}} + \vec{r_{1}}|} = \frac{\vec{v_{2}} - \vec{v_{1}}}{|\vec{v_{2}} + \vec{v_{1}}|}$

A point moves in a straight line so that its displacement x metre at time t sec is given by $x^{2} = 1 + t^{2}$ . Its acceleration in m/s²at time t sec is

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$\frac{1}{x^{3}}$
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$\frac{1}{x} - \frac{1}{x^{2}}$
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$\frac{1}{x} - \frac{t^{2}}{x^{3}}$
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$\frac{- t}{x^{2}}$

A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 s, the average acceleration during contact is nearly (Take g = 10 m $s^{- 2}$ )

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$1500 \sqrt{2} m s^{- 2} u p w a r d s$
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$1800 m s^{- 2} d o w n w a r d s$
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$1500 \sqrt{5} m s^{- 2} u p w a r d s$
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$1500 \sqrt{2} m s^{- 2} d o w n w a r d s$

A particle starting from rest moves in a circle of radius 'r'. It attains a velocity of $v_{0}$ m/s on completion of n rounds. Its angular acceleration will be:

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$\frac{v_{0}}{n} r a d / s^{2}$
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$\frac{v_{0}^{2}}{2 {πnr}^{2}} r a d / s^{2}$
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$\frac{v_{o}^{2}}{4 {πnr}^{2}} r a d / s^{2}$
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$\frac{v_{0}^{2}}{4 πnr} r a d / s^{2}$

Two bullets are fired horizontally and simultaneously towards each other from rooftops of two buildings (building being 100 m apart and being of the same height of 200 m) with the same velocity of 25 m/s. When and where will the two bullets collide? (g = 10 m/ $s^{2}$ )?

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after 2s at a height of 180 m
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after 2s at a height of 20 m
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after 4s at a height of 120 m
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they will not collide.

The position x of a particle moving along the x-axis varies with time t as x = $20 t - 5 t^{2}$ , where x is in meter and t is in second. The particle reverses its direction of motion at:

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x = 40 m
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x = 10 m
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x = 20 m
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x = 30 m

A body starting from rest moves with uniform acceleration on a horizontal surface. The body covers 3 consecutive equal distances from the beginning in time $t_{1}, t_{2} and t_{3}$ seconds. The ratio of $t_{1} : t_{2} : t_{3}$ is

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1:2:3
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$1 : \sqrt{2} : \sqrt{3}$
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$1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})$
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$\sqrt{3} : \sqrt{2} : 1$

A man starts from point 'P' and goes 20 m due East, then 5 m due North, then 35 m due West, and finally 25 m due South. His displacement from point 'P' is:

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25 m at $\tan^{- 1} (\frac{4}{3})$ West of South
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85 m in South-West
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25 m at $\tan^{- 1} (\frac{3}{4})$ West of South
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15 m at $\tan^{- 1} (\frac{3}{4})$ West of South

An aircraft is flying horizontally at a height of 1 km from the ground with a speed of 200 m/s. An anti-craft missile launcher kept at a point on the ground which is in the vertical plane of the motion of aircraft. The muzzle velocity of the missile is 600 m/s. The missile is fired at a time when the aircraft was vertically above the anti-craft missile launcher. At what angle with horizontal should it be fired so as to hit the aircraft?

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$\sin^{- 1} (\frac{2}{6})$
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$\cos^{- 1} (\frac{1}{3})$
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$\tan^{- 1} (3)$
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Just vertically upward

A man can throw a stone to a maximum height 'h'. The greatest horizontal distance up to which he can throw the stone is

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h
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2h
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$\frac{h}{2}$
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4h

An oblique projectile is projected with a speed u. It takes time t, to reach the maximum height and time t, to come back to the ground. Air resistance is not neglected, then $\frac{t_{1}}{t_{2}}$ is

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> 1
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< 1
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= 1
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Depends on the angle of projection

The speed of a projectile projected from level ground at its maximum height is found to be half of its speed of projection (u). Its maximum height is:

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$\frac{3 u^{2}}{8 g}$
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$\frac{3 u^{2}}{2 g}$
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$\frac{\sqrt{3} u^{2}}{2 g}$
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$\frac{3 u^{2}}{4 g}$

A particle of mass m is projected with a velocity at an angle with the horizontal into a uniform gravitational field. The slope of the trajectory varies with horizontal distance x as:

A projectile thrown at an angle 30° with horizontal from the level ground reaches a maximum height of 20 m. What will be the maximum height if it is thrown at an angle 60° with the same speed?

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60 m
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50 m
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20 m
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30 m

A body moving in a uniform circular motion with speed v, the magnitude of change in its velocity after it rotates by an angle 120° is

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2v
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$\sqrt{3}$ v
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v
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$\frac{v \sqrt{3}}{2}$

A car moves on a circular path such that its speed is given by v = Kt, where K =constant and t is time. Also given: radius of the circular path is r. The net acceleration of the car at time t will be

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$\sqrt{K^{2} + {(\frac{K^{2} t^{2}}{r})}^{2}}$
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2K
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K
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$\sqrt{K^{2} + K^{2} t^{2}}$

An object of mass 10 kg is projected from level ground with speed 40 m/s at angle 30° with horizontal. The rate of change of momentum of object [in SI units] 1 second after the projection is [neglect air resistance]

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100
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50
0%

25
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75

A particle is projected horizontally with speed 20 m/s from a cliff of height 20 m. The magnitude of the velocity of particle when it reaches the ground

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20 m/s
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40 m/s
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20 $\sqrt{2}$ m/s
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Zero

A projectile is thrown with an initial velocity of 20 m/s at an angle of 60° with horizontal, then the angle of the velocity of the projectile with horizontal after time 0.732 sec is:

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45°
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30°
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0°
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15°

If $a_{r} and a_{t}$ represent radial and tangential accelerations, then the motion of particle will be uniformly circular for :

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$a_{r}$ = 0, $a_{t}$ = 0
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$a_{r}$ = 0, $a_{t}$ $\neq$ 0
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$a_{r}$ $\neq$ 0, $a_{t}$ = 0
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$a_{r}$ $\neq$ 0, $a_{t}$ $\neq$ 0

A helicopter flies from a city A to B. The line joining A and B is along North-South direction and its length is 100 km. The speed of the helicopter is kept 100 km/h and the wind blows from West to East with a speed of 60 km/h. The time taken by the helicopter is

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0.75 h
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1.0 h
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1.25 h
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1.33 h

In the uniform circular motion, which of the following physical quantity remains constant?

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Angular velocity
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Angular momentum
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Angular speed
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All of these

Four particles lie initially at the corners of a square of side length L. All the particles start to move with speed v. A moves towards B, B moves towards C, C moves towards D and D moves towards A. The distance covered by a particle till they meet is

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$\frac{L}{\sqrt{5}}$
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L
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$\sqrt{2} L$
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2L

A boat can move with a maximum speed of 10 m/s in still water. If the speed of river water is 5 m/s, then in how much minimum time the boat can cross the river of width 500 m?

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$\frac{100}{\sqrt{5}}$ s
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50 s
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100 s
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150 s

A particle is thrown at an angle of projection $θ$ = 45° with speed u. The average velocity of the particle during its ground to ground flight is

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$\sqrt{2} u$
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$\frac{u}{\sqrt{2}}$
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$\frac{u}{2}$
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0

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Practice Physics Quiz Questions and Answers

0:0:1

Practice Physics Quiz Questions and Answers

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