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PHXI06:WORK ENERGY AND POWER

355230 A mass $m_{1}$ moves with a great velocity. It strikes another mass $m_{2}$ at rest in a head on collision. It comes back along its path with low speed, after collision. Then

1

m_{1} > m_{2}

2

m_{1} < m_{2}

3

m_{1} = m_{2}

4

m_{1} \geq m_{2}

Explanation:

$v_{1} = \frac{(m_{1} - m_{2}) u_{1}}{m_{1} + m_{2}}$ As $v_{1}$ is negative and less than $u_{1}$ , therefore, $m_{1} < m_{2} .$

PHXI06:WORK ENERGY AND POWER

355231 A particle of mass $m$ collides with another stationary particle of mass $M$ . If the particle $m$ stops just after collision, the coefficient of restitution for collision is equal to

1

\frac{m}{M}

2 1

3

\frac{m}{M + m}

4

\frac{M - m}{M + m}

Explanation:

As net horizontal force acting on the system is zero, hence momentum must remain conserved.
Let $u$ is the velocity of $m$ and $V$ is the velocity of $M$ after collision
$\Rightarrow V = \frac{m u}{M}$
As per definition,
$e = \frac{(V_{2} - V_{1})}{(u_{1} - u_{2})} = \frac{V - 0}{u - 0} = \frac{V}{u} = \frac{m}{M}$

PHXI06:WORK ENERGY AND POWER

355232 A ball falling freely from a height of $4.9 m s^{- 1}$ hits a horizontal surface. If $e = 3 / 4$ , then the ball will hit the surface second time after

1

0.5 s

2

1.5 s

3

3.5 s

4

3.4 s

Explanation:

From equation of motion,
$v^{2} = u^{2} + 2 g h$
where, $v$ is final velocity, $u$ is initial velocity and $h$ is height,
Given, $h = 4.9 m, g = 9.8 m s^{- 2}$
and $u = 0$
So, $v = \sqrt{2 g h}$
$= \sqrt{2 \times 9.8 \times 4.9} = 9.8 m s^{- 1}$
Coefficient of restitution $(e)$ of an object is a unique dfractional value representing the ratio of relative velocities before and after an impact.
$∴ v^{'} = \frac{3}{4} \times 9.8$
Time taken from first bounce to second bounce
$t = \frac{2 v^{'}}{g} \Rightarrow t = 2 \times \frac{3}{4} \times 9.8 \times \frac{1}{9.8} = 1.5 s$

PHXI06:WORK ENERGY AND POWER

355233 A sphere $A$ of mass m moving with a velocity hits another stationary sphere $B$ of same mass. If the ratio of the velocities of the spheres after collision is $\frac{v_{A}}{v_{B}} = \frac{1 - e}{1 + e}$ where $e$ is the coefficient of restitution, what is the initial velocity of sphere $A$ with which it strikes?

1

v_{A} - v_{B}

2

v_{A} + v_{B}

3

\frac{v_{A} + v_{B}}{2}

4

v_{B} - v_{A}

Explanation:

Given that $\frac{v_{A}}{v_{B}} = \frac{1 - e}{1 + e}$
$\frac{v_{A}}{v_{B}} - 1 = \frac{1 - e}{1 + e} - 1$
$\Rightarrow v_{B} - v_{A} = \frac{2 e}{1 + e} \cdot v_{B} (1)$
$B u t e = \frac{v_{B} - v_{A}}{u} (2)$
$\Rightarrow u = \frac{2 v_{B}}{(1 + e)} \Rightarrow u + e u = 2 v_{B}$
$\Rightarrow u + (v_{B} - v_{A}) = 2 v_{B} \Rightarrow u = v_{A} + v_{B}$

PHXI06:WORK ENERGY AND POWER

355230 A mass $m_{1}$ moves with a great velocity. It strikes another mass $m_{2}$ at rest in a head on collision. It comes back along its path with low speed, after collision. Then

1

m_{1} > m_{2}

2

m_{1} < m_{2}

3

m_{1} = m_{2}

4

m_{1} \geq m_{2}

Explanation:

$v_{1} = \frac{(m_{1} - m_{2}) u_{1}}{m_{1} + m_{2}}$ As $v_{1}$ is negative and less than $u_{1}$ , therefore, $m_{1} < m_{2} .$

PHXI06:WORK ENERGY AND POWER

355231 A particle of mass $m$ collides with another stationary particle of mass $M$ . If the particle $m$ stops just after collision, the coefficient of restitution for collision is equal to

1

\frac{m}{M}

2 1

3

\frac{m}{M + m}

4

\frac{M - m}{M + m}

Explanation:

As net horizontal force acting on the system is zero, hence momentum must remain conserved.
Let $u$ is the velocity of $m$ and $V$ is the velocity of $M$ after collision
$\Rightarrow V = \frac{m u}{M}$
As per definition,
$e = \frac{(V_{2} - V_{1})}{(u_{1} - u_{2})} = \frac{V - 0}{u - 0} = \frac{V}{u} = \frac{m}{M}$

PHXI06:WORK ENERGY AND POWER

355232 A ball falling freely from a height of $4.9 m s^{- 1}$ hits a horizontal surface. If $e = 3 / 4$ , then the ball will hit the surface second time after

1

0.5 s

2

1.5 s

3

3.5 s

4

3.4 s

Explanation:

From equation of motion,
$v^{2} = u^{2} + 2 g h$
where, $v$ is final velocity, $u$ is initial velocity and $h$ is height,
Given, $h = 4.9 m, g = 9.8 m s^{- 2}$
and $u = 0$
So, $v = \sqrt{2 g h}$
$= \sqrt{2 \times 9.8 \times 4.9} = 9.8 m s^{- 1}$
Coefficient of restitution $(e)$ of an object is a unique dfractional value representing the ratio of relative velocities before and after an impact.
$∴ v^{'} = \frac{3}{4} \times 9.8$
Time taken from first bounce to second bounce
$t = \frac{2 v^{'}}{g} \Rightarrow t = 2 \times \frac{3}{4} \times 9.8 \times \frac{1}{9.8} = 1.5 s$

PHXI06:WORK ENERGY AND POWER

355233 A sphere $A$ of mass m moving with a velocity hits another stationary sphere $B$ of same mass. If the ratio of the velocities of the spheres after collision is $\frac{v_{A}}{v_{B}} = \frac{1 - e}{1 + e}$ where $e$ is the coefficient of restitution, what is the initial velocity of sphere $A$ with which it strikes?

1

v_{A} - v_{B}

2

v_{A} + v_{B}

3

\frac{v_{A} + v_{B}}{2}

4

v_{B} - v_{A}

Explanation:

Given that $\frac{v_{A}}{v_{B}} = \frac{1 - e}{1 + e}$
$\frac{v_{A}}{v_{B}} - 1 = \frac{1 - e}{1 + e} - 1$
$\Rightarrow v_{B} - v_{A} = \frac{2 e}{1 + e} \cdot v_{B} (1)$
$B u t e = \frac{v_{B} - v_{A}}{u} (2)$
$\Rightarrow u = \frac{2 v_{B}}{(1 + e)} \Rightarrow u + e u = 2 v_{B}$
$\Rightarrow u + (v_{B} - v_{A}) = 2 v_{B} \Rightarrow u = v_{A} + v_{B}$

PHXI06:WORK ENERGY AND POWER

355230 A mass $m_{1}$ moves with a great velocity. It strikes another mass $m_{2}$ at rest in a head on collision. It comes back along its path with low speed, after collision. Then

1

m_{1} > m_{2}

2

m_{1} < m_{2}

3

m_{1} = m_{2}

4

m_{1} \geq m_{2}

Explanation:

$v_{1} = \frac{(m_{1} - m_{2}) u_{1}}{m_{1} + m_{2}}$ As $v_{1}$ is negative and less than $u_{1}$ , therefore, $m_{1} < m_{2} .$

PHXI06:WORK ENERGY AND POWER

355231 A particle of mass $m$ collides with another stationary particle of mass $M$ . If the particle $m$ stops just after collision, the coefficient of restitution for collision is equal to

1

\frac{m}{M}

2 1

3

\frac{m}{M + m}

4

\frac{M - m}{M + m}

Explanation:

As net horizontal force acting on the system is zero, hence momentum must remain conserved.
Let $u$ is the velocity of $m$ and $V$ is the velocity of $M$ after collision
$\Rightarrow V = \frac{m u}{M}$
As per definition,
$e = \frac{(V_{2} - V_{1})}{(u_{1} - u_{2})} = \frac{V - 0}{u - 0} = \frac{V}{u} = \frac{m}{M}$

PHXI06:WORK ENERGY AND POWER

355232 A ball falling freely from a height of $4.9 m s^{- 1}$ hits a horizontal surface. If $e = 3 / 4$ , then the ball will hit the surface second time after

1

0.5 s

2

1.5 s

3

3.5 s

4

3.4 s

Explanation:

From equation of motion,
$v^{2} = u^{2} + 2 g h$
where, $v$ is final velocity, $u$ is initial velocity and $h$ is height,
Given, $h = 4.9 m, g = 9.8 m s^{- 2}$
and $u = 0$
So, $v = \sqrt{2 g h}$
$= \sqrt{2 \times 9.8 \times 4.9} = 9.8 m s^{- 1}$
Coefficient of restitution $(e)$ of an object is a unique dfractional value representing the ratio of relative velocities before and after an impact.
$∴ v^{'} = \frac{3}{4} \times 9.8$
Time taken from first bounce to second bounce
$t = \frac{2 v^{'}}{g} \Rightarrow t = 2 \times \frac{3}{4} \times 9.8 \times \frac{1}{9.8} = 1.5 s$

PHXI06:WORK ENERGY AND POWER

355233 A sphere $A$ of mass m moving with a velocity hits another stationary sphere $B$ of same mass. If the ratio of the velocities of the spheres after collision is $\frac{v_{A}}{v_{B}} = \frac{1 - e}{1 + e}$ where $e$ is the coefficient of restitution, what is the initial velocity of sphere $A$ with which it strikes?

1

v_{A} - v_{B}

2

v_{A} + v_{B}

3

\frac{v_{A} + v_{B}}{2}

4

v_{B} - v_{A}

Explanation:

Given that $\frac{v_{A}}{v_{B}} = \frac{1 - e}{1 + e}$
$\frac{v_{A}}{v_{B}} - 1 = \frac{1 - e}{1 + e} - 1$
$\Rightarrow v_{B} - v_{A} = \frac{2 e}{1 + e} \cdot v_{B} (1)$
$B u t e = \frac{v_{B} - v_{A}}{u} (2)$
$\Rightarrow u = \frac{2 v_{B}}{(1 + e)} \Rightarrow u + e u = 2 v_{B}$
$\Rightarrow u + (v_{B} - v_{A}) = 2 v_{B} \Rightarrow u = v_{A} + v_{B}$

PHXI06:WORK ENERGY AND POWER

355230 A mass $m_{1}$ moves with a great velocity. It strikes another mass $m_{2}$ at rest in a head on collision. It comes back along its path with low speed, after collision. Then

1

m_{1} > m_{2}

2

m_{1} < m_{2}

3

m_{1} = m_{2}

4

m_{1} \geq m_{2}

Explanation:

$v_{1} = \frac{(m_{1} - m_{2}) u_{1}}{m_{1} + m_{2}}$ As $v_{1}$ is negative and less than $u_{1}$ , therefore, $m_{1} < m_{2} .$

PHXI06:WORK ENERGY AND POWER

355231 A particle of mass $m$ collides with another stationary particle of mass $M$ . If the particle $m$ stops just after collision, the coefficient of restitution for collision is equal to

1

\frac{m}{M}

2 1

3

\frac{m}{M + m}

4

\frac{M - m}{M + m}

Explanation:

As net horizontal force acting on the system is zero, hence momentum must remain conserved.
Let $u$ is the velocity of $m$ and $V$ is the velocity of $M$ after collision
$\Rightarrow V = \frac{m u}{M}$
As per definition,
$e = \frac{(V_{2} - V_{1})}{(u_{1} - u_{2})} = \frac{V - 0}{u - 0} = \frac{V}{u} = \frac{m}{M}$

PHXI06:WORK ENERGY AND POWER

355232 A ball falling freely from a height of $4.9 m s^{- 1}$ hits a horizontal surface. If $e = 3 / 4$ , then the ball will hit the surface second time after

1

0.5 s

2

1.5 s

3

3.5 s

4

3.4 s

Explanation:

From equation of motion,
$v^{2} = u^{2} + 2 g h$
where, $v$ is final velocity, $u$ is initial velocity and $h$ is height,
Given, $h = 4.9 m, g = 9.8 m s^{- 2}$
and $u = 0$
So, $v = \sqrt{2 g h}$
$= \sqrt{2 \times 9.8 \times 4.9} = 9.8 m s^{- 1}$
Coefficient of restitution $(e)$ of an object is a unique dfractional value representing the ratio of relative velocities before and after an impact.
$∴ v^{'} = \frac{3}{4} \times 9.8$
Time taken from first bounce to second bounce
$t = \frac{2 v^{'}}{g} \Rightarrow t = 2 \times \frac{3}{4} \times 9.8 \times \frac{1}{9.8} = 1.5 s$

PHXI06:WORK ENERGY AND POWER

355233 A sphere $A$ of mass m moving with a velocity hits another stationary sphere $B$ of same mass. If the ratio of the velocities of the spheres after collision is $\frac{v_{A}}{v_{B}} = \frac{1 - e}{1 + e}$ where $e$ is the coefficient of restitution, what is the initial velocity of sphere $A$ with which it strikes?

1

v_{A} - v_{B}

2

v_{A} + v_{B}

3

\frac{v_{A} + v_{B}}{2}

4

v_{B} - v_{A}

Explanation:

Given that $\frac{v_{A}}{v_{B}} = \frac{1 - e}{1 + e}$
$\frac{v_{A}}{v_{B}} - 1 = \frac{1 - e}{1 + e} - 1$
$\Rightarrow v_{B} - v_{A} = \frac{2 e}{1 + e} \cdot v_{B} (1)$
$B u t e = \frac{v_{B} - v_{A}}{u} (2)$
$\Rightarrow u = \frac{2 v_{B}}{(1 + e)} \Rightarrow u + e u = 2 v_{B}$
$\Rightarrow u + (v_{B} - v_{A}) = 2 v_{B} \Rightarrow u = v_{A} + v_{B}$

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