Mass $A$ is released from rest at the top of a frictionless inclined plane $18 m$ long and reaches the bottom $3 s$ later. At the instant when $A$ is released, a second mass $B$ is projected upwards along the plate from the bottom with a certain initial velocity. Mass $B$ travels a distance up the plane, stops and returns to the bottom so that it arrives simultaneously with $A$ . The two masses do not collide. Initial velocity of $B$ is

Question

Verified by Toppr

Answer 1

Here, for $A, 18 = 0 \times 3 + \frac{1}{2} a \times 3^{2}$ or $a = 4 m s^{- 2}$
For $B$ , time taken to move up is given by , $t_{1} = u / a$ ( $∴$ the relation $v = u + a t$ here becomes $0 = u - a t_{1}$ ).
Distance moved up is given by the relation $0 = u^{2} - 2 a S$
i.e $S = u^{2} / 2 a$

For coming down the inclined plane for $S = \frac{1}{2} a t_{2}^{2}$

and $t_{2} = \frac{2 S}{a}$

or $t_{2} = \frac{2 u ^{2}}{a . 2 a} = \frac{u}{a}$

But $\frac{u}{a} = t_{1}$ , then

$t_{1} + t_{2} = 3$ or $\frac{2 u}{a} = 3$ or

$u = \frac{3 a}{2}$ or

$u = \frac{3}{2} \times 4 = 6 m s^{- 1}$

Answer 2

Here, for

A, 18 = 0 \times 3 + \frac{1}{2} a \times 3^{2}

or

a = 4 m s^{- 2}

For

B

, time taken to move up is given by ,

t_{1} = u / a

(

∴

the relation

v = u + a t

here becomes

0 = u - a t_{1}

).
Distance moved up is given by the relation

0 = u^{2} - 2 a S

i.e

S = u^{2} / 2 a

For coming down the inclined plane for

S = \frac{1}{2} a t_{2}^{2}

and

t_{2} = \frac{2 S}{a}

or

t_{2} = \frac{2 u ^{2}}{a . 2 a} = \frac{u}{a}

But

\frac{u}{a} = t_{1}

, then

t_{1} + t_{2} = 3

or

\frac{2 u}{a} = 3

or

u = \frac{3 a}{2}

or

u = \frac{3}{2} \times 4 = 6 m s^{- 1}

$4 m s^{- 1}$

$5 m s^{- 1}$

$6 m s^{- 1}$

$7 m s^{- 1}$

A body is allowed to slide from the top along a smooth inclined plane of length 5m at an angle of inclination $3 0^{o}$ . If $g = 10 m s^{- 2}$ , time taken by the body to reach the bottom of the plane is

An object of mass $m$ is released from rest form the top of a smooth inclined plane of height $h$ . Its speed at the bottom of the plane is proportional to

Starting from rest at the top of an inclined plane a body reaches the bottom of the inclined plane in 4 second. In what time dies the body cover one-fourth the distance starting from rest at the top?

Revise with Concepts

Velocity - Time Relation in Uniform Accelerated Motion

Position - Time Relation in Uniform Accelerated Motion

Position - Velocity Relation in Uniform Accelerated Motion

Derivation of Equations of Motion in One Dimension Using Calculus

Displacement in Nth Second

4ms−1

5ms−1

6ms−1

7ms−1

A body is allowed to slide from the top along a smooth inclined plane of length 5m at an angle of inclination 30o. If g=10ms−2, time taken by the body to reach the bottom of the plane is

An object of mass m is released from rest form the top of a smooth inclined plane of height h. Its speed at the bottom of the plane is proportional to

Starting from rest at the top of an inclined plane a body reaches the bottom of the inclined plane in 4 second. In what time dies the body cover one-fourth the distance starting from rest at the top?

Revise with Concepts

Velocity - Time Relation in Uniform Accelerated Motion

Position - Time Relation in Uniform Accelerated Motion

Position - Velocity Relation in Uniform Accelerated Motion

Derivation of Equations of Motion in One Dimension Using Calculus

Displacement in Nth Second

$4 m s^{- 1}$

$5 m s^{- 1}$

$6 m s^{- 1}$

$7 m s^{- 1}$

A body is allowed to slide from the top along a smooth inclined plane of length 5m at an angle of inclination $3 0^{o}$ . If $g = 10 m s^{- 2}$ , time taken by the body to reach the bottom of the plane is

An object of mass $m$ is released from rest form the top of a smooth inclined plane of height $h$ . Its speed at the bottom of the plane is proportional to