QBManager

Work, Energy and Power

149175 A ball is projected vertically down with an initial velocity from a height $10 m$ onto a horizontal floor. During the impact it loses $20 %$ of the energy and rebounds to the same height. What is the initial velocity of its projection?
(Use $g = 10 m / s^{2}$ )

1

5 \sqrt{2} m / s

2

10 \sqrt{3} m / s

3

5 \sqrt{3} m / s

4

10 \sqrt{2} m / s

Explanation:

A Let a ball is projected vertically downward with velocity $v$ from height $h$
Total energy at point $A = \frac{1}{2} {mv}^{2} + mgh$

During collision loss of energy is $20 %$ and the ball rises up to same height.
$(\frac{1}{2} {mv}^{2} + mgh) 80 % = mgh$
$(\frac{1}{2} {mv}^{2} + mgh) \frac{4}{5} = mgh$
$\frac{v^{2}}{2} + gh = \frac{5 gh}{4}$
$\frac{v^{2}}{2} = \frac{5 gh}{4} - gh$
$\frac{v^{2}}{2} = gh (\frac{5}{4} - 1)$
$v^{2} = \frac{gh}{4} \times 2$
$v^{2} = \frac{gh}{2}$
$v = \sqrt{\frac{10 \times 10}{2}}$
$v = 5 \sqrt{2} m / s$

TS EAMCET 28.09.2020

Work, Energy and Power

149176 A tennis ball hits the floor with a speed $v$ at an angle $θ$ with the normal to the floor. If the collision is inelastic and the coefficient of restitution is $ε$ , what will be the angle of reflection?

1

\tan^{- 1} (\frac{\tan θ}{ε})

2

\sin^{- 1} (\frac{\sin θ}{ε})

3

θ ε

4

θ \frac{2 ε}{ε + 1}

Explanation:

A Velocity of approach $=$ normal component of velocity.

Along horizontal direction momentum is conserved.
$v \sin ϕ = u \sin θ$
Along the vertical
$ε = \frac{v \cos ϕ}{u \cos θ} = \frac{\sin θ}{\sin ϕ} \times \frac{\cos ϕ}{\cos θ} [∵ \frac{v}{u} = \frac{\sin θ}{\sin ϕ}]$
$\Rightarrow ε = \frac{\tan θ}{\tan ϕ}$
$\Rightarrow \tan ϕ = \frac{\tan θ}{ε}$
$\Rightarrow= \tan^{- 1} (\frac{\tan θ}{ε})$

WB JEE 2020

Work, Energy and Power

149177 A block of mass ' $m$ ' collides with another stationary block of mass ' $2 m$ '. The lighter block comes to rest after collision. If the velocity of first block is ' $u$ ', then the value of coefficient of restitution is

1 0.8

2 0.4

3 0.5

4 0.6

Explanation:

C Given,
$m_{1} = m, m_{2} = 2 m$
$u_{1} = u, u_{2} = 0$
$v_{1} = 0, v_{2} = v^{'}$
$Before collision After collision$

By law of conservation of momentum:-
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$mu = m (0) + 2 m v^{'}$
$v^{'} = \frac{u}{2}$
$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$
$e = \frac{v^{'} - 0}{u - 0} = \frac{v^{'}}{u}$
Putting the value of $v^{'}$ in equation (ii), we get -
$e = \frac{u}{2 u}$
$= \frac{1}{2} = 0.5$

MHT-CET 2020

Work, Energy and Power

149178 A block of mass ' $m$ ' moving on a frictionless surface at speed ' $V$ ' collides elastically with a block of same mass, initially at rest, Now the first block moves at an angle ' $θ$ ' with its initial direction and has speed ' $V_{1}$ '. The speed of the second block after collision is

1

\sqrt{V^{2} - V_{1}^{2}}

2

\sqrt{V_{1}^{2} - V^{2}}

3

\sqrt{V^{2} + V_{1}^{2}}

4

\sqrt{V - V_{1}}

Explanation:

A Given, $m_{1} = m_{2} = m$
$u_{1} = V, u_{2} = 0$

at rest
From conservation of K.E-
K.E before the collision and K.E. after the collision are equal.
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} V_{1}^{2} + \frac{1}{2} m_{2} V_{2}^{2}$
$\frac{1}{2} {mV}^{2} + 0 = \frac{1}{2} {mV}_{1}^{2} + \frac{1}{2} {mV}_{2}^{2}$
$V^{2} = V_{1}^{2} + V_{2}^{2}$
$V_{2}^{2} = V^{2} - V_{1}^{2}$
$V_{2} = \sqrt{V^{2} - V_{1}^{2}}$

MHT-CET 2020

Work, Energy and Power

149175 A ball is projected vertically down with an initial velocity from a height $10 m$ onto a horizontal floor. During the impact it loses $20 %$ of the energy and rebounds to the same height. What is the initial velocity of its projection?
(Use $g = 10 m / s^{2}$ )

1

5 \sqrt{2} m / s

2

10 \sqrt{3} m / s

3

5 \sqrt{3} m / s

4

10 \sqrt{2} m / s

Explanation:

A Let a ball is projected vertically downward with velocity $v$ from height $h$
Total energy at point $A = \frac{1}{2} {mv}^{2} + mgh$

During collision loss of energy is $20 %$ and the ball rises up to same height.
$(\frac{1}{2} {mv}^{2} + mgh) 80 % = mgh$
$(\frac{1}{2} {mv}^{2} + mgh) \frac{4}{5} = mgh$
$\frac{v^{2}}{2} + gh = \frac{5 gh}{4}$
$\frac{v^{2}}{2} = \frac{5 gh}{4} - gh$
$\frac{v^{2}}{2} = gh (\frac{5}{4} - 1)$
$v^{2} = \frac{gh}{4} \times 2$
$v^{2} = \frac{gh}{2}$
$v = \sqrt{\frac{10 \times 10}{2}}$
$v = 5 \sqrt{2} m / s$

TS EAMCET 28.09.2020

Work, Energy and Power

149176 A tennis ball hits the floor with a speed $v$ at an angle $θ$ with the normal to the floor. If the collision is inelastic and the coefficient of restitution is $ε$ , what will be the angle of reflection?

1

\tan^{- 1} (\frac{\tan θ}{ε})

2

\sin^{- 1} (\frac{\sin θ}{ε})

3

θ ε

4

θ \frac{2 ε}{ε + 1}

Explanation:

A Velocity of approach $=$ normal component of velocity.

Along horizontal direction momentum is conserved.
$v \sin ϕ = u \sin θ$
Along the vertical
$ε = \frac{v \cos ϕ}{u \cos θ} = \frac{\sin θ}{\sin ϕ} \times \frac{\cos ϕ}{\cos θ} [∵ \frac{v}{u} = \frac{\sin θ}{\sin ϕ}]$
$\Rightarrow ε = \frac{\tan θ}{\tan ϕ}$
$\Rightarrow \tan ϕ = \frac{\tan θ}{ε}$
$\Rightarrow= \tan^{- 1} (\frac{\tan θ}{ε})$

WB JEE 2020

Work, Energy and Power

149177 A block of mass ' $m$ ' collides with another stationary block of mass ' $2 m$ '. The lighter block comes to rest after collision. If the velocity of first block is ' $u$ ', then the value of coefficient of restitution is

1 0.8

2 0.4

3 0.5

4 0.6

Explanation:

C Given,
$m_{1} = m, m_{2} = 2 m$
$u_{1} = u, u_{2} = 0$
$v_{1} = 0, v_{2} = v^{'}$
$Before collision After collision$

By law of conservation of momentum:-
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$mu = m (0) + 2 m v^{'}$
$v^{'} = \frac{u}{2}$
$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$
$e = \frac{v^{'} - 0}{u - 0} = \frac{v^{'}}{u}$
Putting the value of $v^{'}$ in equation (ii), we get -
$e = \frac{u}{2 u}$
$= \frac{1}{2} = 0.5$

MHT-CET 2020

Work, Energy and Power

149178 A block of mass ' $m$ ' moving on a frictionless surface at speed ' $V$ ' collides elastically with a block of same mass, initially at rest, Now the first block moves at an angle ' $θ$ ' with its initial direction and has speed ' $V_{1}$ '. The speed of the second block after collision is

1

\sqrt{V^{2} - V_{1}^{2}}

2

\sqrt{V_{1}^{2} - V^{2}}

3

\sqrt{V^{2} + V_{1}^{2}}

4

\sqrt{V - V_{1}}

Explanation:

A Given, $m_{1} = m_{2} = m$
$u_{1} = V, u_{2} = 0$

at rest
From conservation of K.E-
K.E before the collision and K.E. after the collision are equal.
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} V_{1}^{2} + \frac{1}{2} m_{2} V_{2}^{2}$
$\frac{1}{2} {mV}^{2} + 0 = \frac{1}{2} {mV}_{1}^{2} + \frac{1}{2} {mV}_{2}^{2}$
$V^{2} = V_{1}^{2} + V_{2}^{2}$
$V_{2}^{2} = V^{2} - V_{1}^{2}$
$V_{2} = \sqrt{V^{2} - V_{1}^{2}}$

MHT-CET 2020

Work, Energy and Power

149175 A ball is projected vertically down with an initial velocity from a height $10 m$ onto a horizontal floor. During the impact it loses $20 %$ of the energy and rebounds to the same height. What is the initial velocity of its projection?
(Use $g = 10 m / s^{2}$ )

1

5 \sqrt{2} m / s

2

10 \sqrt{3} m / s

3

5 \sqrt{3} m / s

4

10 \sqrt{2} m / s

Explanation:

A Let a ball is projected vertically downward with velocity $v$ from height $h$
Total energy at point $A = \frac{1}{2} {mv}^{2} + mgh$

During collision loss of energy is $20 %$ and the ball rises up to same height.
$(\frac{1}{2} {mv}^{2} + mgh) 80 % = mgh$
$(\frac{1}{2} {mv}^{2} + mgh) \frac{4}{5} = mgh$
$\frac{v^{2}}{2} + gh = \frac{5 gh}{4}$
$\frac{v^{2}}{2} = \frac{5 gh}{4} - gh$
$\frac{v^{2}}{2} = gh (\frac{5}{4} - 1)$
$v^{2} = \frac{gh}{4} \times 2$
$v^{2} = \frac{gh}{2}$
$v = \sqrt{\frac{10 \times 10}{2}}$
$v = 5 \sqrt{2} m / s$

TS EAMCET 28.09.2020

Work, Energy and Power

149176 A tennis ball hits the floor with a speed $v$ at an angle $θ$ with the normal to the floor. If the collision is inelastic and the coefficient of restitution is $ε$ , what will be the angle of reflection?

1

\tan^{- 1} (\frac{\tan θ}{ε})

2

\sin^{- 1} (\frac{\sin θ}{ε})

3

θ ε

4

θ \frac{2 ε}{ε + 1}

Explanation:

A Velocity of approach $=$ normal component of velocity.

Along horizontal direction momentum is conserved.
$v \sin ϕ = u \sin θ$
Along the vertical
$ε = \frac{v \cos ϕ}{u \cos θ} = \frac{\sin θ}{\sin ϕ} \times \frac{\cos ϕ}{\cos θ} [∵ \frac{v}{u} = \frac{\sin θ}{\sin ϕ}]$
$\Rightarrow ε = \frac{\tan θ}{\tan ϕ}$
$\Rightarrow \tan ϕ = \frac{\tan θ}{ε}$
$\Rightarrow= \tan^{- 1} (\frac{\tan θ}{ε})$

WB JEE 2020

Work, Energy and Power

149177 A block of mass ' $m$ ' collides with another stationary block of mass ' $2 m$ '. The lighter block comes to rest after collision. If the velocity of first block is ' $u$ ', then the value of coefficient of restitution is

1 0.8

2 0.4

3 0.5

4 0.6

Explanation:

C Given,
$m_{1} = m, m_{2} = 2 m$
$u_{1} = u, u_{2} = 0$
$v_{1} = 0, v_{2} = v^{'}$
$Before collision After collision$

By law of conservation of momentum:-
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$mu = m (0) + 2 m v^{'}$
$v^{'} = \frac{u}{2}$
$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$
$e = \frac{v^{'} - 0}{u - 0} = \frac{v^{'}}{u}$
Putting the value of $v^{'}$ in equation (ii), we get -
$e = \frac{u}{2 u}$
$= \frac{1}{2} = 0.5$

MHT-CET 2020

Work, Energy and Power

149178 A block of mass ' $m$ ' moving on a frictionless surface at speed ' $V$ ' collides elastically with a block of same mass, initially at rest, Now the first block moves at an angle ' $θ$ ' with its initial direction and has speed ' $V_{1}$ '. The speed of the second block after collision is

1

\sqrt{V^{2} - V_{1}^{2}}

2

\sqrt{V_{1}^{2} - V^{2}}

3

\sqrt{V^{2} + V_{1}^{2}}

4

\sqrt{V - V_{1}}

Explanation:

A Given, $m_{1} = m_{2} = m$
$u_{1} = V, u_{2} = 0$

at rest
From conservation of K.E-
K.E before the collision and K.E. after the collision are equal.
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} V_{1}^{2} + \frac{1}{2} m_{2} V_{2}^{2}$
$\frac{1}{2} {mV}^{2} + 0 = \frac{1}{2} {mV}_{1}^{2} + \frac{1}{2} {mV}_{2}^{2}$
$V^{2} = V_{1}^{2} + V_{2}^{2}$
$V_{2}^{2} = V^{2} - V_{1}^{2}$
$V_{2} = \sqrt{V^{2} - V_{1}^{2}}$

MHT-CET 2020

Work, Energy and Power

149175 A ball is projected vertically down with an initial velocity from a height $10 m$ onto a horizontal floor. During the impact it loses $20 %$ of the energy and rebounds to the same height. What is the initial velocity of its projection?
(Use $g = 10 m / s^{2}$ )

1

5 \sqrt{2} m / s

2

10 \sqrt{3} m / s

3

5 \sqrt{3} m / s

4

10 \sqrt{2} m / s

Explanation:

A Let a ball is projected vertically downward with velocity $v$ from height $h$
Total energy at point $A = \frac{1}{2} {mv}^{2} + mgh$

During collision loss of energy is $20 %$ and the ball rises up to same height.
$(\frac{1}{2} {mv}^{2} + mgh) 80 % = mgh$
$(\frac{1}{2} {mv}^{2} + mgh) \frac{4}{5} = mgh$
$\frac{v^{2}}{2} + gh = \frac{5 gh}{4}$
$\frac{v^{2}}{2} = \frac{5 gh}{4} - gh$
$\frac{v^{2}}{2} = gh (\frac{5}{4} - 1)$
$v^{2} = \frac{gh}{4} \times 2$
$v^{2} = \frac{gh}{2}$
$v = \sqrt{\frac{10 \times 10}{2}}$
$v = 5 \sqrt{2} m / s$

TS EAMCET 28.09.2020

Work, Energy and Power

149176 A tennis ball hits the floor with a speed $v$ at an angle $θ$ with the normal to the floor. If the collision is inelastic and the coefficient of restitution is $ε$ , what will be the angle of reflection?

1

\tan^{- 1} (\frac{\tan θ}{ε})

2

\sin^{- 1} (\frac{\sin θ}{ε})

3

θ ε

4

θ \frac{2 ε}{ε + 1}

Explanation:

A Velocity of approach $=$ normal component of velocity.

Along horizontal direction momentum is conserved.
$v \sin ϕ = u \sin θ$
Along the vertical
$ε = \frac{v \cos ϕ}{u \cos θ} = \frac{\sin θ}{\sin ϕ} \times \frac{\cos ϕ}{\cos θ} [∵ \frac{v}{u} = \frac{\sin θ}{\sin ϕ}]$
$\Rightarrow ε = \frac{\tan θ}{\tan ϕ}$
$\Rightarrow \tan ϕ = \frac{\tan θ}{ε}$
$\Rightarrow= \tan^{- 1} (\frac{\tan θ}{ε})$

WB JEE 2020

Work, Energy and Power

149177 A block of mass ' $m$ ' collides with another stationary block of mass ' $2 m$ '. The lighter block comes to rest after collision. If the velocity of first block is ' $u$ ', then the value of coefficient of restitution is

1 0.8

2 0.4

3 0.5

4 0.6

Explanation:

C Given,
$m_{1} = m, m_{2} = 2 m$
$u_{1} = u, u_{2} = 0$
$v_{1} = 0, v_{2} = v^{'}$
$Before collision After collision$

By law of conservation of momentum:-
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$mu = m (0) + 2 m v^{'}$
$v^{'} = \frac{u}{2}$
$e = \frac{v_{2} - v_{1}}{u_{1} - u_{2}}$
$e = \frac{v^{'} - 0}{u - 0} = \frac{v^{'}}{u}$
Putting the value of $v^{'}$ in equation (ii), we get -
$e = \frac{u}{2 u}$
$= \frac{1}{2} = 0.5$

MHT-CET 2020

Work, Energy and Power

149178 A block of mass ' $m$ ' moving on a frictionless surface at speed ' $V$ ' collides elastically with a block of same mass, initially at rest, Now the first block moves at an angle ' $θ$ ' with its initial direction and has speed ' $V_{1}$ '. The speed of the second block after collision is

1

\sqrt{V^{2} - V_{1}^{2}}

2

\sqrt{V_{1}^{2} - V^{2}}

3

\sqrt{V^{2} + V_{1}^{2}}

4

\sqrt{V - V_{1}}

Explanation:

A Given, $m_{1} = m_{2} = m$
$u_{1} = V, u_{2} = 0$

at rest
From conservation of K.E-
K.E before the collision and K.E. after the collision are equal.
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} V_{1}^{2} + \frac{1}{2} m_{2} V_{2}^{2}$
$\frac{1}{2} {mV}^{2} + 0 = \frac{1}{2} {mV}_{1}^{2} + \frac{1}{2} {mV}_{2}^{2}$
$V^{2} = V_{1}^{2} + V_{2}^{2}$
$V_{2}^{2} = V^{2} - V_{1}^{2}$
$V_{2} = \sqrt{V^{2} - V_{1}^{2}}$

MHT-CET 2020

Question Bank Series

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