Two particles, 1 and 2 , move with constant velocities v 1 → and v 2 → . At the initial moment their radius vectors are equal to r 1 → and r 2 → . How must these four vectors be interrelated for the particles to collide?

Here, the situation is shown in the following diagram, Two particles are initially at points C and D at time t = 0 . Let us assume that the two particles collide at the point A at time t = t . Let the position vector of point A be r 3 → . Now, C A → = Displacement of the particle 1 in time t = t = v 1 → t and D A → = Displacement of the particle 2 in time t = t = v 2 → t Now, using the triangle law of vector addition for each particle, we get, r 3 → = r 1 → + v 1 → t and r 3 → = r 2 → + v 2 → t Hence, we get, r 1 → + v 1 → t = r 2 → + v 2 → t ⇒ r 1 → - r 2 → = v 2 → - v 1 → t . . . ( 1 ) ∴ t = r 1 → - r 2 → v 2 → - v 1 → . . . ( 2 ) Putting equation ( 2 ) in ( 1 ) , we get, r 1 → - r 2 → = v 2 → - v 1 → r 1 → - r 2 → v 2 → - v 1 → . This is the required relation.

Two particles move in a uniform gravitational field with an acceleration g . At the initial moment the particles were located at one point and moved with velocities v 1 → = 3 . 0 m s - 1 and v 2 → = 4 . 0 m s - 1 horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular. (Take g = 9 . 8 m s - 2 )

Let the velocities of the particles (say v 1 ' → and v 2 ' → ) becomes mutually perpendicular after a time t . So, v 1 ' → = v 1 → + g → t , v 1 ' → = v 2 → + g → t . . . 1 As v 1 ' → ⊥ v 2 ' → so, v 1 ' → · v 2 ' → = 0 or v 1 → + g → t · v 2 → + g → t = 0 or - v 1 v 2 + g 2 t 2 = 0 , (As v → 1 and v → 2 are opposite to each other) Hence, t = v 1 v 2 g . . . 3 Since both the particles start from the same point, their initial relative position r → 0 12 = r → 0 1 - r → 0 2 = 0 . Both the particles moves under Earth's gravitational field with same acceleration due to gravity. Therefore, their relative acceleration, w → 12 = g → - g → = 0 . Their initial relative velocity, v → 0 12 = v → 1 + v → 2 = v 1 i ^ - - v 2 i ^ = v 1 + v 2 i ^ Now, from the second equation of uniformly accelerated motion, r → 12 = r → 0 12 + v → 0 12 t + 1 2 w 12 → t 2 We get, r → 12 = v → 0 12 t , ( Here, w → 12 = 0 and r → 0 ( 12 ) = 0 Hence, the sought distance is, r → 12 = v 1 + v 2 g v 1 v 2 , as v → 0 ( 12 ) = v 1 + v 2 r → 12 = 3 + 4 9 . 8 3 × 4 = 2 . 47 m

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IMPORTANT

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A point moves rectilinearly in one direction, as shown below.

The distance $s$ traversed by the point as a function of the time $t$ . Using the plot, find,
$(a)$ the average velocity of the point during the time of motion,
$(b)$ the maximum velocity,
$(c)$ the time moment $t_{0}$ at which the instantaneous velocity is equal to the mean velocity averaged over the first $t_{0}$ seconds.

Important Questions on PHYSICAL FUNDAMENTALS OF MECHANICS

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JEE Main

IMPORTANT

Problems in General Physics>Chapter 1 - PHYSICAL FUNDAMENTALS OF MECHANICS>KINEMATICS>Q 1.5.

Two particles,

1

and

2

, move with constant velocities

\vec{v_{1}}

and

\vec{v_{2}}

. At the initial moment their radius vectors are equal to

\vec{r_{1}}

and

\vec{r_{2}}

. How must these four vectors be interrelated for the particles to collide?

HARD

JEE Main

IMPORTANT

Problems in General Physics>Chapter 1 - PHYSICAL FUNDAMENTALS OF MECHANICS>KINEMATICS>Q 1.6.

A ship moves along the equator to the east with velocity

v_{0} = 30 km h^{- 1}

. The southeastern wind blows at an angle

φ = 60 °

to the equator with velocity

v = 15 km h^{- 1}

. Find the wind velocity

v^{'}

relative to the ship and the angle

θ

between the equator and the wind direction in the reference frame fixed to the ship.

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IMPORTANT

Problems in General Physics>Chapter 1 - PHYSICAL FUNDAMENTALS OF MECHANICS>KINEMATICS>Q 1.7.

Two swimmers leave a point $A$ on one bank of the river to reach point $B$ lying right across on the other bank. One of them crosses the river along the straight line $A B$ while the other swims at right angles to the stream and then walks the distance that he has been carried away by the stream to get to point $B$ . What was the velocity $u$ of his walking if both swimmers reached the destination simultaneously? The stream velocity $v_{0} = 2.0 km h^{- 1}$ and the velocity $v^{'}$ of each swimmer with respect to water equals $2.5 km h^{- 1}$ .

HARD

JEE Main

IMPORTANT

Problems in General Physics>Chapter 1 - PHYSICAL FUNDAMENTALS OF MECHANICS>KINEMATICS>Q 1.8.

Two boats,

A

and

B

, move away from a buoy anchored at the middle of a river along the mutually perpendicular straight lines- the boat

A

along the river, and the boat

B

across the river. Having moved off an equal distance from the buoy the boats returned. Find the ratio of times of motion of boats

\frac{t_{A}}{t_{B}}

, if the velocity of each boat with respect to water is

η = 1.2

times greater than the stream velocity.

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JEE Main

IMPORTANT

Problems in General Physics>Chapter 1 - PHYSICAL FUNDAMENTALS OF MECHANICS>KINEMATICS>Q 1.9.

A boat moves relative to water with a velocity which is

η = 2.0

times less than the river flow velocity. At what angle to the stream direction must the boat move to minimize drifting?

HARD

JEE Main

IMPORTANT

Problems in General Physics>Chapter 1 - PHYSICAL FUNDAMENTALS OF MECHANICS>KINEMATICS>Q 1.10.

Two bodies were thrown simultaneously from the same point -one, straight up, and the other, at an angle of

θ = 60 °

to the horizontal. The initial velocity of each body is equal to

v_{0} = 25 m s^{- 1}

. Neglecting the air drag, find the distance between the bodies

t = 1.70

s

later.

HARD

JEE Main

IMPORTANT

Problems in General Physics>Chapter 1 - PHYSICAL FUNDAMENTALS OF MECHANICS>KINEMATICS>Q 1.11.

Two particles move in a uniform gravitational field with an acceleration

g

. At the initial moment the particles were located at one point and moved with velocities

\vec{v_{1}} = 3.0 m s^{- 1}

and

\vec{v_{2}} = 4.0 m s^{- 1}

horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular. (Take

g = 9.8 m s^{- 2}

)

HARD

JEE Main

IMPORTANT

Problems in General Physics>Chapter 1 - PHYSICAL FUNDAMENTALS OF MECHANICS>KINEMATICS>Q 1.12.

Three points are located at the vertices of an equilateral triangle whose side equals

a

. They all start moving simultaneously with velocity

v

constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?

Important Questions on PHYSICAL FUNDAMENTALS OF MECHANICS

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