QBManager

PHXI04:MOTION IN A PLANE

362003 A particle is projected with speed $u$ that makes an angle $θ$ with horizontal, the time at which the velocity of the projectile is perpendicular to the initial velocity is

1

\frac{u}{g \cos θ}

2

\frac{u}{g \tan θ}

3

\frac{u}{g \sin θ}

4

\frac{u \sin θ}{g}

Explanation:

The initial velocity vector is
$\vec{u} = u \cos θ \hat{i} + u \sin θ \hat{j}$
The velocity vector after time $t$ is
$\vec{v} = u \cos θ \hat{i} + (u \sin θ - g t) \hat{j}$
When $\vec{v}$ is perpendicular to $\vec{u}$
$\vec{v} . \vec{u} = 0$
$u^{2} \cos^{2} + u^{2} \sin^{2} θ - u \sin θ g t = 0$
$t = \frac{u}{g \sin θ}$

PHXI04:MOTION IN A PLANE

362004 A projectile of mass $m$ is thrown with a velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The change in momentum from departure to arrival along vertical direction, is

1

2 m v

2

\sqrt{2} m v

3

m v

4

\frac{4 R}{g}

Explanation:

Change in momentum is the product of force and time.
$Δ p = m g \times \frac{2 \sin θ}{g}$ $(∵ F = \frac{Δ p}{Δ t})$
$= 2 m v \sin θ = 2 m v \sin 45^{\circ}$
$= \frac{2 m v}{\sqrt{2}} = \sqrt{2} m v$

NCERT Exemplar

PHXI04:MOTION IN A PLANE

362005 A projectile is fired making an angle $2 θ$ with horizontal with velocity $4 m s^{- 1}$ . At some instant it makes an angle $θ$ , then its velocity is :

1

4 (2 \cos θ - \sec θ)

2

4 \cos θ

3

4 (\sec θ + \cos θ)

4

2 (\sec θ + 4 \cos θ)

Explanation:

Let $u$ be the velocity when projected with angle $2 θ$ , then equating the horizontal velocities in both the cases, we get
$v \cos θ = u \cos 2 θ \Rightarrow v = \frac{u \cos 2 θ}{\cos θ}$
Where, $\sec θ = \frac{1}{\cos θ}$
$∴ v = u \cos 2 θ \sec θ$
Using $\cos 2 θ = 2 \cos^{2} θ - 1$ , and $u = 4 m s^{- 1}$
$v = 4 (2 \cos^{2} θ - 1) \sec θ \Rightarrow v = 4 (2 \cos θ - \sec θ)$

PHXI04:MOTION IN A PLANE

362006 The path of one projectile as seen from another projectile is a

1 Straight line

2 Parabola

3 Hyperbola

4 Circles

Explanation:

$y_{1} = u_{1} \sin α_{1} t - \frac{1}{2} g t^{2} a n d x_{1} = u_{1} \cos α_{1} . t$
$y_{2} = u_{2} \sin α_{2} t - \frac{1}{2} g t^{2} a n d x_{2} = u_{2} \cos α_{2} . t$
$∴ \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{u_{2} \sin α_{2} - u_{1} \sin α_{1}}{u_{2} \cos α_{2} - u_{1} \cos α_{1}} = c o n s t a n t$

PHXI04:MOTION IN A PLANE

362008 The velocity of a projectile at the initial point $A$ is $(4 \hat{i} + 5 \hat{j}) m / s$ Its velocity (in $m$ / $s$ ) at point $B$ is
supporting img

1

4 \hat{i} + 5 \hat{j}

2

- 4 \hat{i} - 5 \hat{j}

3

- 4 \hat{i} + 5 \hat{j}

4

4 \hat{i} - 5 \hat{j}

PHXI04:MOTION IN A PLANE

362003 A particle is projected with speed $u$ that makes an angle $θ$ with horizontal, the time at which the velocity of the projectile is perpendicular to the initial velocity is

1

\frac{u}{g \cos θ}

2

\frac{u}{g \tan θ}

3

\frac{u}{g \sin θ}

4

\frac{u \sin θ}{g}

Explanation:

The initial velocity vector is
$\vec{u} = u \cos θ \hat{i} + u \sin θ \hat{j}$
The velocity vector after time $t$ is
$\vec{v} = u \cos θ \hat{i} + (u \sin θ - g t) \hat{j}$
When $\vec{v}$ is perpendicular to $\vec{u}$
$\vec{v} . \vec{u} = 0$
$u^{2} \cos^{2} + u^{2} \sin^{2} θ - u \sin θ g t = 0$
$t = \frac{u}{g \sin θ}$

PHXI04:MOTION IN A PLANE

362004 A projectile of mass $m$ is thrown with a velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The change in momentum from departure to arrival along vertical direction, is

1

2 m v

2

\sqrt{2} m v

3

m v

4

\frac{4 R}{g}

Explanation:

Change in momentum is the product of force and time.
$Δ p = m g \times \frac{2 \sin θ}{g}$ $(∵ F = \frac{Δ p}{Δ t})$
$= 2 m v \sin θ = 2 m v \sin 45^{\circ}$
$= \frac{2 m v}{\sqrt{2}} = \sqrt{2} m v$

NCERT Exemplar

PHXI04:MOTION IN A PLANE

362005 A projectile is fired making an angle $2 θ$ with horizontal with velocity $4 m s^{- 1}$ . At some instant it makes an angle $θ$ , then its velocity is :

1

4 (2 \cos θ - \sec θ)

2

4 \cos θ

3

4 (\sec θ + \cos θ)

4

2 (\sec θ + 4 \cos θ)

Explanation:

Let $u$ be the velocity when projected with angle $2 θ$ , then equating the horizontal velocities in both the cases, we get
$v \cos θ = u \cos 2 θ \Rightarrow v = \frac{u \cos 2 θ}{\cos θ}$
Where, $\sec θ = \frac{1}{\cos θ}$
$∴ v = u \cos 2 θ \sec θ$
Using $\cos 2 θ = 2 \cos^{2} θ - 1$ , and $u = 4 m s^{- 1}$
$v = 4 (2 \cos^{2} θ - 1) \sec θ \Rightarrow v = 4 (2 \cos θ - \sec θ)$

PHXI04:MOTION IN A PLANE

362006 The path of one projectile as seen from another projectile is a

1 Straight line

2 Parabola

3 Hyperbola

4 Circles

Explanation:

$y_{1} = u_{1} \sin α_{1} t - \frac{1}{2} g t^{2} a n d x_{1} = u_{1} \cos α_{1} . t$
$y_{2} = u_{2} \sin α_{2} t - \frac{1}{2} g t^{2} a n d x_{2} = u_{2} \cos α_{2} . t$
$∴ \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{u_{2} \sin α_{2} - u_{1} \sin α_{1}}{u_{2} \cos α_{2} - u_{1} \cos α_{1}} = c o n s t a n t$

PHXI04:MOTION IN A PLANE

362008 The velocity of a projectile at the initial point $A$ is $(4 \hat{i} + 5 \hat{j}) m / s$ Its velocity (in $m$ / $s$ ) at point $B$ is
supporting img

1

4 \hat{i} + 5 \hat{j}

2

- 4 \hat{i} - 5 \hat{j}

3

- 4 \hat{i} + 5 \hat{j}

4

4 \hat{i} - 5 \hat{j}

PHXI04:MOTION IN A PLANE

362003 A particle is projected with speed $u$ that makes an angle $θ$ with horizontal, the time at which the velocity of the projectile is perpendicular to the initial velocity is

1

\frac{u}{g \cos θ}

2

\frac{u}{g \tan θ}

3

\frac{u}{g \sin θ}

4

\frac{u \sin θ}{g}

Explanation:

The initial velocity vector is
$\vec{u} = u \cos θ \hat{i} + u \sin θ \hat{j}$
The velocity vector after time $t$ is
$\vec{v} = u \cos θ \hat{i} + (u \sin θ - g t) \hat{j}$
When $\vec{v}$ is perpendicular to $\vec{u}$
$\vec{v} . \vec{u} = 0$
$u^{2} \cos^{2} + u^{2} \sin^{2} θ - u \sin θ g t = 0$
$t = \frac{u}{g \sin θ}$

PHXI04:MOTION IN A PLANE

362004 A projectile of mass $m$ is thrown with a velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The change in momentum from departure to arrival along vertical direction, is

1

2 m v

2

\sqrt{2} m v

3

m v

4

\frac{4 R}{g}

Explanation:

Change in momentum is the product of force and time.
$Δ p = m g \times \frac{2 \sin θ}{g}$ $(∵ F = \frac{Δ p}{Δ t})$
$= 2 m v \sin θ = 2 m v \sin 45^{\circ}$
$= \frac{2 m v}{\sqrt{2}} = \sqrt{2} m v$

NCERT Exemplar

PHXI04:MOTION IN A PLANE

362005 A projectile is fired making an angle $2 θ$ with horizontal with velocity $4 m s^{- 1}$ . At some instant it makes an angle $θ$ , then its velocity is :

1

4 (2 \cos θ - \sec θ)

2

4 \cos θ

3

4 (\sec θ + \cos θ)

4

2 (\sec θ + 4 \cos θ)

Explanation:

Let $u$ be the velocity when projected with angle $2 θ$ , then equating the horizontal velocities in both the cases, we get
$v \cos θ = u \cos 2 θ \Rightarrow v = \frac{u \cos 2 θ}{\cos θ}$
Where, $\sec θ = \frac{1}{\cos θ}$
$∴ v = u \cos 2 θ \sec θ$
Using $\cos 2 θ = 2 \cos^{2} θ - 1$ , and $u = 4 m s^{- 1}$
$v = 4 (2 \cos^{2} θ - 1) \sec θ \Rightarrow v = 4 (2 \cos θ - \sec θ)$

PHXI04:MOTION IN A PLANE

362006 The path of one projectile as seen from another projectile is a

1 Straight line

2 Parabola

3 Hyperbola

4 Circles

Explanation:

$y_{1} = u_{1} \sin α_{1} t - \frac{1}{2} g t^{2} a n d x_{1} = u_{1} \cos α_{1} . t$
$y_{2} = u_{2} \sin α_{2} t - \frac{1}{2} g t^{2} a n d x_{2} = u_{2} \cos α_{2} . t$
$∴ \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{u_{2} \sin α_{2} - u_{1} \sin α_{1}}{u_{2} \cos α_{2} - u_{1} \cos α_{1}} = c o n s t a n t$

PHXI04:MOTION IN A PLANE

362008 The velocity of a projectile at the initial point $A$ is $(4 \hat{i} + 5 \hat{j}) m / s$ Its velocity (in $m$ / $s$ ) at point $B$ is
supporting img

1

4 \hat{i} + 5 \hat{j}

2

- 4 \hat{i} - 5 \hat{j}

3

- 4 \hat{i} + 5 \hat{j}

4

4 \hat{i} - 5 \hat{j}

PHXI04:MOTION IN A PLANE

362003 A particle is projected with speed $u$ that makes an angle $θ$ with horizontal, the time at which the velocity of the projectile is perpendicular to the initial velocity is

1

\frac{u}{g \cos θ}

2

\frac{u}{g \tan θ}

3

\frac{u}{g \sin θ}

4

\frac{u \sin θ}{g}

Explanation:

The initial velocity vector is
$\vec{u} = u \cos θ \hat{i} + u \sin θ \hat{j}$
The velocity vector after time $t$ is
$\vec{v} = u \cos θ \hat{i} + (u \sin θ - g t) \hat{j}$
When $\vec{v}$ is perpendicular to $\vec{u}$
$\vec{v} . \vec{u} = 0$
$u^{2} \cos^{2} + u^{2} \sin^{2} θ - u \sin θ g t = 0$
$t = \frac{u}{g \sin θ}$

PHXI04:MOTION IN A PLANE

362004 A projectile of mass $m$ is thrown with a velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The change in momentum from departure to arrival along vertical direction, is

1

2 m v

2

\sqrt{2} m v

3

m v

4

\frac{4 R}{g}

Explanation:

Change in momentum is the product of force and time.
$Δ p = m g \times \frac{2 \sin θ}{g}$ $(∵ F = \frac{Δ p}{Δ t})$
$= 2 m v \sin θ = 2 m v \sin 45^{\circ}$
$= \frac{2 m v}{\sqrt{2}} = \sqrt{2} m v$

NCERT Exemplar

PHXI04:MOTION IN A PLANE

362005 A projectile is fired making an angle $2 θ$ with horizontal with velocity $4 m s^{- 1}$ . At some instant it makes an angle $θ$ , then its velocity is :

1

4 (2 \cos θ - \sec θ)

2

4 \cos θ

3

4 (\sec θ + \cos θ)

4

2 (\sec θ + 4 \cos θ)

Explanation:

Let $u$ be the velocity when projected with angle $2 θ$ , then equating the horizontal velocities in both the cases, we get
$v \cos θ = u \cos 2 θ \Rightarrow v = \frac{u \cos 2 θ}{\cos θ}$
Where, $\sec θ = \frac{1}{\cos θ}$
$∴ v = u \cos 2 θ \sec θ$
Using $\cos 2 θ = 2 \cos^{2} θ - 1$ , and $u = 4 m s^{- 1}$
$v = 4 (2 \cos^{2} θ - 1) \sec θ \Rightarrow v = 4 (2 \cos θ - \sec θ)$

PHXI04:MOTION IN A PLANE

362006 The path of one projectile as seen from another projectile is a

1 Straight line

2 Parabola

3 Hyperbola

4 Circles

Explanation:

$y_{1} = u_{1} \sin α_{1} t - \frac{1}{2} g t^{2} a n d x_{1} = u_{1} \cos α_{1} . t$
$y_{2} = u_{2} \sin α_{2} t - \frac{1}{2} g t^{2} a n d x_{2} = u_{2} \cos α_{2} . t$
$∴ \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{u_{2} \sin α_{2} - u_{1} \sin α_{1}}{u_{2} \cos α_{2} - u_{1} \cos α_{1}} = c o n s t a n t$

PHXI04:MOTION IN A PLANE

362008 The velocity of a projectile at the initial point $A$ is $(4 \hat{i} + 5 \hat{j}) m / s$ Its velocity (in $m$ / $s$ ) at point $B$ is
supporting img

1

4 \hat{i} + 5 \hat{j}

2

- 4 \hat{i} - 5 \hat{j}

3

- 4 \hat{i} + 5 \hat{j}

4

4 \hat{i} - 5 \hat{j}

PHXI04:MOTION IN A PLANE

362003 A particle is projected with speed $u$ that makes an angle $θ$ with horizontal, the time at which the velocity of the projectile is perpendicular to the initial velocity is

1

\frac{u}{g \cos θ}

2

\frac{u}{g \tan θ}

3

\frac{u}{g \sin θ}

4

\frac{u \sin θ}{g}

Explanation:

The initial velocity vector is
$\vec{u} = u \cos θ \hat{i} + u \sin θ \hat{j}$
The velocity vector after time $t$ is
$\vec{v} = u \cos θ \hat{i} + (u \sin θ - g t) \hat{j}$
When $\vec{v}$ is perpendicular to $\vec{u}$
$\vec{v} . \vec{u} = 0$
$u^{2} \cos^{2} + u^{2} \sin^{2} θ - u \sin θ g t = 0$
$t = \frac{u}{g \sin θ}$

PHXI04:MOTION IN A PLANE

362004 A projectile of mass $m$ is thrown with a velocity $v$ making an angle of $45^{\circ}$ with the horizontal. The change in momentum from departure to arrival along vertical direction, is

1

2 m v

2

\sqrt{2} m v

3

m v

4

\frac{4 R}{g}

Explanation:

Change in momentum is the product of force and time.
$Δ p = m g \times \frac{2 \sin θ}{g}$ $(∵ F = \frac{Δ p}{Δ t})$
$= 2 m v \sin θ = 2 m v \sin 45^{\circ}$
$= \frac{2 m v}{\sqrt{2}} = \sqrt{2} m v$

NCERT Exemplar

PHXI04:MOTION IN A PLANE

362005 A projectile is fired making an angle $2 θ$ with horizontal with velocity $4 m s^{- 1}$ . At some instant it makes an angle $θ$ , then its velocity is :

1

4 (2 \cos θ - \sec θ)

2

4 \cos θ

3

4 (\sec θ + \cos θ)

4

2 (\sec θ + 4 \cos θ)

Explanation:

Let $u$ be the velocity when projected with angle $2 θ$ , then equating the horizontal velocities in both the cases, we get
$v \cos θ = u \cos 2 θ \Rightarrow v = \frac{u \cos 2 θ}{\cos θ}$
Where, $\sec θ = \frac{1}{\cos θ}$
$∴ v = u \cos 2 θ \sec θ$
Using $\cos 2 θ = 2 \cos^{2} θ - 1$ , and $u = 4 m s^{- 1}$
$v = 4 (2 \cos^{2} θ - 1) \sec θ \Rightarrow v = 4 (2 \cos θ - \sec θ)$

PHXI04:MOTION IN A PLANE

362006 The path of one projectile as seen from another projectile is a

1 Straight line

2 Parabola

3 Hyperbola

4 Circles

Explanation:

$y_{1} = u_{1} \sin α_{1} t - \frac{1}{2} g t^{2} a n d x_{1} = u_{1} \cos α_{1} . t$
$y_{2} = u_{2} \sin α_{2} t - \frac{1}{2} g t^{2} a n d x_{2} = u_{2} \cos α_{2} . t$
$∴ \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{u_{2} \sin α_{2} - u_{1} \sin α_{1}}{u_{2} \cos α_{2} - u_{1} \cos α_{1}} = c o n s t a n t$

PHXI04:MOTION IN A PLANE

362008 The velocity of a projectile at the initial point $A$ is $(4 \hat{i} + 5 \hat{j}) m / s$ Its velocity (in $m$ / $s$ ) at point $B$ is
supporting img

1

4 \hat{i} + 5 \hat{j}

2

- 4 \hat{i} - 5 \hat{j}

3

- 4 \hat{i} + 5 \hat{j}

4

4 \hat{i} - 5 \hat{j}

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: