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PHXI06:WORK ENERGY AND POWER

355213 A bullet of mass $M$ hits a block of mass $M^{'}$ . The energy transfer is maximum, when

1

M^{'} = 2 M

2

M^{'} = M

3

M^{'} >> M

4

M^{'} << M

Explanation:

If $M = M^{'}$ , then bullet will transfer whole of its velocity ( $100 %$ of its $K E$ ) to block and will itself come to rest.

PHXI06:WORK ENERGY AND POWER

355214 Consider the following statements
( $A$ ) Linear momentum of a system of particles is zero
( $B$ ) Kinetic energy of a system of particles is zero then

1

A

does not imply

B

&

B

does not imply

A

2

A

implies

B

and

B

does not imply

A

3

A

does not imply

B

but

B

implies

A

4

A

implies

B

and

B

implies

A

PHXI06:WORK ENERGY AND POWER

355215 When two spheres of equal masses undergo glancing elastic collision with one of them at rest, after collision they will move:

1 In the same direction

2 Opposite to one another

3 At right angle to each other

4 Together

PHXI06:WORK ENERGY AND POWER

355216 A ball is projected with initial velocity $u$ at an angle $θ$ to the horizontal. Then horizontal displacement covered by ball as it collides third time to the ground be, if coefficient of restitution is $e$ :
supporting img

1

e \frac{u^{2} \sin 2 θ}{g}

2

(1 + e) \frac{u^{2} \sin 2 θ}{g}

3

(1 + e + e^{2}) \frac{u^{2} \sin 2 θ}{g}

4

(1 - e) \frac{u^{2} \sin 2 θ}{g}

Explanation:

After each collision vertical velocity becomes e times whereas horizontal velocity remain same. So after each collision time of flight becomes e times of previous one. So that horizontal displacement
$= R + e R + e^{2} R$
$= (1 + e + e^{2}) \frac{u^{2} \sin 2 θ}{g}$

PHXI06:WORK ENERGY AND POWER

355213 A bullet of mass $M$ hits a block of mass $M^{'}$ . The energy transfer is maximum, when

1

M^{'} = 2 M

2

M^{'} = M

3

M^{'} >> M

4

M^{'} << M

Explanation:

If $M = M^{'}$ , then bullet will transfer whole of its velocity ( $100 %$ of its $K E$ ) to block and will itself come to rest.

PHXI06:WORK ENERGY AND POWER

355214 Consider the following statements
( $A$ ) Linear momentum of a system of particles is zero
( $B$ ) Kinetic energy of a system of particles is zero then

1

A

does not imply

B

&

B

does not imply

A

2

A

implies

B

and

B

does not imply

A

3

A

does not imply

B

but

B

implies

A

4

A

implies

B

and

B

implies

A

PHXI06:WORK ENERGY AND POWER

355215 When two spheres of equal masses undergo glancing elastic collision with one of them at rest, after collision they will move:

1 In the same direction

2 Opposite to one another

3 At right angle to each other

4 Together

PHXI06:WORK ENERGY AND POWER

355216 A ball is projected with initial velocity $u$ at an angle $θ$ to the horizontal. Then horizontal displacement covered by ball as it collides third time to the ground be, if coefficient of restitution is $e$ :
supporting img

1

e \frac{u^{2} \sin 2 θ}{g}

2

(1 + e) \frac{u^{2} \sin 2 θ}{g}

3

(1 + e + e^{2}) \frac{u^{2} \sin 2 θ}{g}

4

(1 - e) \frac{u^{2} \sin 2 θ}{g}

Explanation:

After each collision vertical velocity becomes e times whereas horizontal velocity remain same. So after each collision time of flight becomes e times of previous one. So that horizontal displacement
$= R + e R + e^{2} R$
$= (1 + e + e^{2}) \frac{u^{2} \sin 2 θ}{g}$

PHXI06:WORK ENERGY AND POWER

355213 A bullet of mass $M$ hits a block of mass $M^{'}$ . The energy transfer is maximum, when

1

M^{'} = 2 M

2

M^{'} = M

3

M^{'} >> M

4

M^{'} << M

Explanation:

If $M = M^{'}$ , then bullet will transfer whole of its velocity ( $100 %$ of its $K E$ ) to block and will itself come to rest.

PHXI06:WORK ENERGY AND POWER

355214 Consider the following statements
( $A$ ) Linear momentum of a system of particles is zero
( $B$ ) Kinetic energy of a system of particles is zero then

1

A

does not imply

B

&

B

does not imply

A

2

A

implies

B

and

B

does not imply

A

3

A

does not imply

B

but

B

implies

A

4

A

implies

B

and

B

implies

A

PHXI06:WORK ENERGY AND POWER

355215 When two spheres of equal masses undergo glancing elastic collision with one of them at rest, after collision they will move:

1 In the same direction

2 Opposite to one another

3 At right angle to each other

4 Together

PHXI06:WORK ENERGY AND POWER

355216 A ball is projected with initial velocity $u$ at an angle $θ$ to the horizontal. Then horizontal displacement covered by ball as it collides third time to the ground be, if coefficient of restitution is $e$ :
supporting img

1

e \frac{u^{2} \sin 2 θ}{g}

2

(1 + e) \frac{u^{2} \sin 2 θ}{g}

3

(1 + e + e^{2}) \frac{u^{2} \sin 2 θ}{g}

4

(1 - e) \frac{u^{2} \sin 2 θ}{g}

Explanation:

After each collision vertical velocity becomes e times whereas horizontal velocity remain same. So after each collision time of flight becomes e times of previous one. So that horizontal displacement
$= R + e R + e^{2} R$
$= (1 + e + e^{2}) \frac{u^{2} \sin 2 θ}{g}$

PHXI06:WORK ENERGY AND POWER

355213 A bullet of mass $M$ hits a block of mass $M^{'}$ . The energy transfer is maximum, when

1

M^{'} = 2 M

2

M^{'} = M

3

M^{'} >> M

4

M^{'} << M

Explanation:

If $M = M^{'}$ , then bullet will transfer whole of its velocity ( $100 %$ of its $K E$ ) to block and will itself come to rest.

PHXI06:WORK ENERGY AND POWER

355214 Consider the following statements
( $A$ ) Linear momentum of a system of particles is zero
( $B$ ) Kinetic energy of a system of particles is zero then

1

A

does not imply

B

&

B

does not imply

A

2

A

implies

B

and

B

does not imply

A

3

A

does not imply

B

but

B

implies

A

4

A

implies

B

and

B

implies

A

PHXI06:WORK ENERGY AND POWER

355215 When two spheres of equal masses undergo glancing elastic collision with one of them at rest, after collision they will move:

1 In the same direction

2 Opposite to one another

3 At right angle to each other

4 Together

PHXI06:WORK ENERGY AND POWER

355216 A ball is projected with initial velocity $u$ at an angle $θ$ to the horizontal. Then horizontal displacement covered by ball as it collides third time to the ground be, if coefficient of restitution is $e$ :
supporting img

1

e \frac{u^{2} \sin 2 θ}{g}

2

(1 + e) \frac{u^{2} \sin 2 θ}{g}

3

(1 + e + e^{2}) \frac{u^{2} \sin 2 θ}{g}

4

(1 - e) \frac{u^{2} \sin 2 θ}{g}

Explanation:

After each collision vertical velocity becomes e times whereas horizontal velocity remain same. So after each collision time of flight becomes e times of previous one. So that horizontal displacement
$= R + e R + e^{2} R$
$= (1 + e + e^{2}) \frac{u^{2} \sin 2 θ}{g}$

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