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PHXI05:LAWS OF MOTION

363259 A rod is hinged about the fixed point $O$ and it can rotate about $O$ . If the rod is in contact with the wedge, then find the relation between $v_{1} & v_{2}$
supporting img

1

v_{1} = v_{2} \sin θ

2

v_{1} = v_{2} \cos θ

3

v_{2} = v_{1} \sin θ

4

v_{1} = v_{2}

Explanation:

The surfaces of rod and wedge that are in contact are as shown below.
$v_{2} = v_{1} \cos (90 - θ) \Rightarrow v_{2} = v_{1} \sin θ$
supporting img

PHXI05:LAWS OF MOTION

363260 If the rod $A$ and wedge $B$ are in contact then find the relation between $v_{A}$ & $v_{B}$
supporting img

1

v_{B} = v_{A} \tan θ

2

v_{B} = v_{A} \cot θ

3

v_{B} = v_{A} \sin θ

4

v_{B} = v_{A} \cos θ

Explanation:

The two surfaces which are in contact are shown below.
$v_{A} \cos (90 - θ) = v_{B} \cos θ$
The velocity components must be same along the normal.
$v_{A} \sin θ = v_{B} \cos θ$
$\Rightarrow v_{B} = v_{A} \tan θ .$
supporting img

PHXI05:LAWS OF MOTION

363261 If the block and the wedge are in contact then find the condition in terms of velocities.
supporting img

1

v_{2} \sin θ = v_{1 y} \cos θ - v_{1 x} \sin θ

2

v_{2} \sin θ + v_{1 y} \cos θ = v_{1 x} \sin θ

3

v_{2} = v_{1 x}

4

N o n e o f t h e s e

Explanation:

The two surfaces which are in contact are shown below.
$v_{2}$ component $=$ net $v_{1}$ component (along normal direction)
$v_{2} \sin θ = v_{1 y} \cos θ - v_{1 x} \sin θ$
supporting img

PHXI05:LAWS OF MOTION

363262 If the two blocks are in contact with each other, then find the relation between $v_{1} & v_{2}$
supporting img

1

v_{1} = v_{2} \sin θ

2

v_{1} = v_{2} \tan θ

3

v_{2} = v_{1} \sin θ

4

v_{1} = v_{2} \cos θ

Explanation:

The two surfaces are as shown below.
$v_{1} \cos θ = v_{2} \cos (90 - θ)$
$v_{1} \cos θ = v_{2} \sin θ$
$v_{1} = v_{2} \tan θ$
supporting img

PHXI05:LAWS OF MOTION

363263 In the arrangement shown in the figure, if the acceleration of $B$ is a then the acceleration of $A$ is
supporting img

1

a \sin α

2

a Tan θ

3

a \cot θ

4

a (\cos α + \sin α \cot θ)

Explanation:

The block $B$ slides along the fixed inclined surface. Acceleration of $B$ makes an angle $90 - α$ w.r.to vertical direction.The surfaces of $A$ and $B$ which are in contact are shown below.
Here $b$ is the acceleration of $A$
From the figure
$x + 90 - θ + 90 - α = 90 \Rightarrow x = (α + θ) - 90$
As the two surfaces are in contact
$a \cos [(α + θ) - 90] = b \cos (90 - θ)$
$a \sin (α + θ) = b \sin θ$
$b = \frac{a}{\sin θ} [\sin α \cos θ + \cos α \sin θ]$
$b = a [\sin α \cot θ + \cos α]$
supporting img

PHXI05:LAWS OF MOTION

363259 A rod is hinged about the fixed point $O$ and it can rotate about $O$ . If the rod is in contact with the wedge, then find the relation between $v_{1} & v_{2}$
supporting img

1

v_{1} = v_{2} \sin θ

2

v_{1} = v_{2} \cos θ

3

v_{2} = v_{1} \sin θ

4

v_{1} = v_{2}

Explanation:

The surfaces of rod and wedge that are in contact are as shown below.
$v_{2} = v_{1} \cos (90 - θ) \Rightarrow v_{2} = v_{1} \sin θ$
supporting img

PHXI05:LAWS OF MOTION

363260 If the rod $A$ and wedge $B$ are in contact then find the relation between $v_{A}$ & $v_{B}$
supporting img

1

v_{B} = v_{A} \tan θ

2

v_{B} = v_{A} \cot θ

3

v_{B} = v_{A} \sin θ

4

v_{B} = v_{A} \cos θ

Explanation:

The two surfaces which are in contact are shown below.
$v_{A} \cos (90 - θ) = v_{B} \cos θ$
The velocity components must be same along the normal.
$v_{A} \sin θ = v_{B} \cos θ$
$\Rightarrow v_{B} = v_{A} \tan θ .$
supporting img

PHXI05:LAWS OF MOTION

363261 If the block and the wedge are in contact then find the condition in terms of velocities.
supporting img

1

v_{2} \sin θ = v_{1 y} \cos θ - v_{1 x} \sin θ

2

v_{2} \sin θ + v_{1 y} \cos θ = v_{1 x} \sin θ

3

v_{2} = v_{1 x}

4

N o n e o f t h e s e

Explanation:

The two surfaces which are in contact are shown below.
$v_{2}$ component $=$ net $v_{1}$ component (along normal direction)
$v_{2} \sin θ = v_{1 y} \cos θ - v_{1 x} \sin θ$
supporting img

PHXI05:LAWS OF MOTION

363262 If the two blocks are in contact with each other, then find the relation between $v_{1} & v_{2}$
supporting img

1

v_{1} = v_{2} \sin θ

2

v_{1} = v_{2} \tan θ

3

v_{2} = v_{1} \sin θ

4

v_{1} = v_{2} \cos θ

Explanation:

The two surfaces are as shown below.
$v_{1} \cos θ = v_{2} \cos (90 - θ)$
$v_{1} \cos θ = v_{2} \sin θ$
$v_{1} = v_{2} \tan θ$
supporting img

PHXI05:LAWS OF MOTION

363263 In the arrangement shown in the figure, if the acceleration of $B$ is a then the acceleration of $A$ is
supporting img

1

a \sin α

2

a Tan θ

3

a \cot θ

4

a (\cos α + \sin α \cot θ)

Explanation:

The block $B$ slides along the fixed inclined surface. Acceleration of $B$ makes an angle $90 - α$ w.r.to vertical direction.The surfaces of $A$ and $B$ which are in contact are shown below.
Here $b$ is the acceleration of $A$
From the figure
$x + 90 - θ + 90 - α = 90 \Rightarrow x = (α + θ) - 90$
As the two surfaces are in contact
$a \cos [(α + θ) - 90] = b \cos (90 - θ)$
$a \sin (α + θ) = b \sin θ$
$b = \frac{a}{\sin θ} [\sin α \cos θ + \cos α \sin θ]$
$b = a [\sin α \cot θ + \cos α]$
supporting img

PHXI05:LAWS OF MOTION

363259 A rod is hinged about the fixed point $O$ and it can rotate about $O$ . If the rod is in contact with the wedge, then find the relation between $v_{1} & v_{2}$
supporting img

1

v_{1} = v_{2} \sin θ

2

v_{1} = v_{2} \cos θ

3

v_{2} = v_{1} \sin θ

4

v_{1} = v_{2}

Explanation:

The surfaces of rod and wedge that are in contact are as shown below.
$v_{2} = v_{1} \cos (90 - θ) \Rightarrow v_{2} = v_{1} \sin θ$
supporting img

PHXI05:LAWS OF MOTION

363260 If the rod $A$ and wedge $B$ are in contact then find the relation between $v_{A}$ & $v_{B}$
supporting img

1

v_{B} = v_{A} \tan θ

2

v_{B} = v_{A} \cot θ

3

v_{B} = v_{A} \sin θ

4

v_{B} = v_{A} \cos θ

Explanation:

The two surfaces which are in contact are shown below.
$v_{A} \cos (90 - θ) = v_{B} \cos θ$
The velocity components must be same along the normal.
$v_{A} \sin θ = v_{B} \cos θ$
$\Rightarrow v_{B} = v_{A} \tan θ .$
supporting img

PHXI05:LAWS OF MOTION

363261 If the block and the wedge are in contact then find the condition in terms of velocities.
supporting img

1

v_{2} \sin θ = v_{1 y} \cos θ - v_{1 x} \sin θ

2

v_{2} \sin θ + v_{1 y} \cos θ = v_{1 x} \sin θ

3

v_{2} = v_{1 x}

4

N o n e o f t h e s e

Explanation:

The two surfaces which are in contact are shown below.
$v_{2}$ component $=$ net $v_{1}$ component (along normal direction)
$v_{2} \sin θ = v_{1 y} \cos θ - v_{1 x} \sin θ$
supporting img

PHXI05:LAWS OF MOTION

363262 If the two blocks are in contact with each other, then find the relation between $v_{1} & v_{2}$
supporting img

1

v_{1} = v_{2} \sin θ

2

v_{1} = v_{2} \tan θ

3

v_{2} = v_{1} \sin θ

4

v_{1} = v_{2} \cos θ

Explanation:

The two surfaces are as shown below.
$v_{1} \cos θ = v_{2} \cos (90 - θ)$
$v_{1} \cos θ = v_{2} \sin θ$
$v_{1} = v_{2} \tan θ$
supporting img

PHXI05:LAWS OF MOTION

363263 In the arrangement shown in the figure, if the acceleration of $B$ is a then the acceleration of $A$ is
supporting img

1

a \sin α

2

a Tan θ

3

a \cot θ

4

a (\cos α + \sin α \cot θ)

Explanation:

The block $B$ slides along the fixed inclined surface. Acceleration of $B$ makes an angle $90 - α$ w.r.to vertical direction.The surfaces of $A$ and $B$ which are in contact are shown below.
Here $b$ is the acceleration of $A$
From the figure
$x + 90 - θ + 90 - α = 90 \Rightarrow x = (α + θ) - 90$
As the two surfaces are in contact
$a \cos [(α + θ) - 90] = b \cos (90 - θ)$
$a \sin (α + θ) = b \sin θ$
$b = \frac{a}{\sin θ} [\sin α \cos θ + \cos α \sin θ]$
$b = a [\sin α \cot θ + \cos α]$
supporting img

PHXI05:LAWS OF MOTION

363259 A rod is hinged about the fixed point $O$ and it can rotate about $O$ . If the rod is in contact with the wedge, then find the relation between $v_{1} & v_{2}$
supporting img

1

v_{1} = v_{2} \sin θ

2

v_{1} = v_{2} \cos θ

3

v_{2} = v_{1} \sin θ

4

v_{1} = v_{2}

Explanation:

The surfaces of rod and wedge that are in contact are as shown below.
$v_{2} = v_{1} \cos (90 - θ) \Rightarrow v_{2} = v_{1} \sin θ$
supporting img

PHXI05:LAWS OF MOTION

363260 If the rod $A$ and wedge $B$ are in contact then find the relation between $v_{A}$ & $v_{B}$
supporting img

1

v_{B} = v_{A} \tan θ

2

v_{B} = v_{A} \cot θ

3

v_{B} = v_{A} \sin θ

4

v_{B} = v_{A} \cos θ

Explanation:

The two surfaces which are in contact are shown below.
$v_{A} \cos (90 - θ) = v_{B} \cos θ$
The velocity components must be same along the normal.
$v_{A} \sin θ = v_{B} \cos θ$
$\Rightarrow v_{B} = v_{A} \tan θ .$
supporting img

PHXI05:LAWS OF MOTION

363261 If the block and the wedge are in contact then find the condition in terms of velocities.
supporting img

1

v_{2} \sin θ = v_{1 y} \cos θ - v_{1 x} \sin θ

2

v_{2} \sin θ + v_{1 y} \cos θ = v_{1 x} \sin θ

3

v_{2} = v_{1 x}

4

N o n e o f t h e s e

Explanation:

The two surfaces which are in contact are shown below.
$v_{2}$ component $=$ net $v_{1}$ component (along normal direction)
$v_{2} \sin θ = v_{1 y} \cos θ - v_{1 x} \sin θ$
supporting img

PHXI05:LAWS OF MOTION

363262 If the two blocks are in contact with each other, then find the relation between $v_{1} & v_{2}$
supporting img

1

v_{1} = v_{2} \sin θ

2

v_{1} = v_{2} \tan θ

3

v_{2} = v_{1} \sin θ

4

v_{1} = v_{2} \cos θ

Explanation:

The two surfaces are as shown below.
$v_{1} \cos θ = v_{2} \cos (90 - θ)$
$v_{1} \cos θ = v_{2} \sin θ$
$v_{1} = v_{2} \tan θ$
supporting img

PHXI05:LAWS OF MOTION

363263 In the arrangement shown in the figure, if the acceleration of $B$ is a then the acceleration of $A$ is
supporting img

1

a \sin α

2

a Tan θ

3

a \cot θ

4

a (\cos α + \sin α \cot θ)

Explanation:

The block $B$ slides along the fixed inclined surface. Acceleration of $B$ makes an angle $90 - α$ w.r.to vertical direction.The surfaces of $A$ and $B$ which are in contact are shown below.
Here $b$ is the acceleration of $A$
From the figure
$x + 90 - θ + 90 - α = 90 \Rightarrow x = (α + θ) - 90$
As the two surfaces are in contact
$a \cos [(α + θ) - 90] = b \cos (90 - θ)$
$a \sin (α + θ) = b \sin θ$
$b = \frac{a}{\sin θ} [\sin α \cos θ + \cos α \sin θ]$
$b = a [\sin α \cot θ + \cos α]$
supporting img

PHXI05:LAWS OF MOTION

363259 A rod is hinged about the fixed point $O$ and it can rotate about $O$ . If the rod is in contact with the wedge, then find the relation between $v_{1} & v_{2}$
supporting img

1

v_{1} = v_{2} \sin θ

2

v_{1} = v_{2} \cos θ

3

v_{2} = v_{1} \sin θ

4

v_{1} = v_{2}

Explanation:

The surfaces of rod and wedge that are in contact are as shown below.
$v_{2} = v_{1} \cos (90 - θ) \Rightarrow v_{2} = v_{1} \sin θ$
supporting img

PHXI05:LAWS OF MOTION

363260 If the rod $A$ and wedge $B$ are in contact then find the relation between $v_{A}$ & $v_{B}$
supporting img

1

v_{B} = v_{A} \tan θ

2

v_{B} = v_{A} \cot θ

3

v_{B} = v_{A} \sin θ

4

v_{B} = v_{A} \cos θ

Explanation:

The two surfaces which are in contact are shown below.
$v_{A} \cos (90 - θ) = v_{B} \cos θ$
The velocity components must be same along the normal.
$v_{A} \sin θ = v_{B} \cos θ$
$\Rightarrow v_{B} = v_{A} \tan θ .$
supporting img

PHXI05:LAWS OF MOTION

363261 If the block and the wedge are in contact then find the condition in terms of velocities.
supporting img

1

v_{2} \sin θ = v_{1 y} \cos θ - v_{1 x} \sin θ

2

v_{2} \sin θ + v_{1 y} \cos θ = v_{1 x} \sin θ

3

v_{2} = v_{1 x}

4

N o n e o f t h e s e

Explanation:

The two surfaces which are in contact are shown below.
$v_{2}$ component $=$ net $v_{1}$ component (along normal direction)
$v_{2} \sin θ = v_{1 y} \cos θ - v_{1 x} \sin θ$
supporting img

PHXI05:LAWS OF MOTION

363262 If the two blocks are in contact with each other, then find the relation between $v_{1} & v_{2}$
supporting img

1

v_{1} = v_{2} \sin θ

2

v_{1} = v_{2} \tan θ

3

v_{2} = v_{1} \sin θ

4

v_{1} = v_{2} \cos θ

Explanation:

The two surfaces are as shown below.
$v_{1} \cos θ = v_{2} \cos (90 - θ)$
$v_{1} \cos θ = v_{2} \sin θ$
$v_{1} = v_{2} \tan θ$
supporting img

PHXI05:LAWS OF MOTION

363263 In the arrangement shown in the figure, if the acceleration of $B$ is a then the acceleration of $A$ is
supporting img

1

a \sin α

2

a Tan θ

3

a \cot θ

4

a (\cos α + \sin α \cot θ)

Explanation:

The block $B$ slides along the fixed inclined surface. Acceleration of $B$ makes an angle $90 - α$ w.r.to vertical direction.The surfaces of $A$ and $B$ which are in contact are shown below.
Here $b$ is the acceleration of $A$
From the figure
$x + 90 - θ + 90 - α = 90 \Rightarrow x = (α + θ) - 90$
As the two surfaces are in contact
$a \cos [(α + θ) - 90] = b \cos (90 - θ)$
$a \sin (α + θ) = b \sin θ$
$b = \frac{a}{\sin θ} [\sin α \cos θ + \cos α \sin θ]$
$b = a [\sin α \cot θ + \cos α]$
supporting img

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