QBManager

Work, Energy and Power

149164 A ball of mass $0.15 kg$ hits the wall with its initial speed of $12 {ms}^{- 1}$ and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is $100 N$ . calculate the time duration of the contact of ball with the wall.

1

0.018 s

2

0.036 s

3

0.009 s

4

0.072 s

Explanation:

B Given,
$Mass of ball (m) = 0.15 kg$
$Initial speed of ball (u) = 12 m / s$
$v = - 12 m / s, F = 100 N, Δ t = ?$
Initial momentum $(p_{i}) = mu = 12 \times 0.15$
$p_{i} = 1.8 kgm / s$
Final momentum $(p_{f}) = mv = - 12 \times 0.15$
$p_{f} = - 1.8 kgm / s$
Change in momentum,
$Δ p = p_{i} - p_{f} = 1.8 - (- 1.8)$
$Δ p = 3.6 kg m / s$
We know that,
Change in momentum $(Δ p) = F \times Δ t$
$Δ t = \frac{Δ P}{F} = \frac{3.6}{100} = 0.036 s$

JEE Main-26.07.2022

Work, Energy and Power

149165 Two objects I and II of mass $0.8 kg$ and $0.6 kg$ collide elastically. If before collision, the object II was at rest and object I was moving. Find the ratio of the final velocity of object $I$ and object II.

1

6 : 7

2

4 : 3

3

1 : 8

4

1 : 16

Explanation:

C Given, $m_{1} = 0.8 kg, m_{2} = 0.6 kg$ ,
$u_{2} = 0$
$e = 1 (∵ elastically collide)$
Law of conservation of momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$0.8 u_{1} + 0 = 0.8 v_{1} + 0.6 v_{2}$
$0.8 v_{1} + 0.6 v_{2} = 0.8 u_{1}$
Newton's experimental law
$v_{2} - v_{1} = e (u_{1} - u_{2}) (∵ e = 1)$
$v_{2} - v_{1} = 1 (u_{1} - 0)$
$u_{1} = v_{2} - v_{1}$
From equation (i), we get
$0.8 v_{1} + 0.6 v_{2} = 0.8 (v_{2} - v_{1})$
$0.8 v_{1} + 0.6 v_{2} = 0.8 v_{2} - 0.8 v_{1}$
$1.6 v_{1} = 0.2 v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{2}{16}$
$v_{1} : v_{2} = 1 : 8$

TS EAMCET 30.07.2022

Work, Energy and Power

149166 Two identical bodies of same mass move with speed $V_{0}$ an $2 V_{0}$ , respectively in directions perpendicular to each other. After some time, the bodies collide and merge with each other. The speed after collision is

1

\frac{V_{0}}{2}

2

\frac{V_{0} \sqrt{5}}{2}

3

\frac{V_{0}}{2 \sqrt{5}}

4

V_{0}

Explanation:

B Given,
Two identical bodies of same mass i.e. $m_{1} = m_{2} = m$ Moving with speed $V_{0}$ and $2 V_{0}$
$∵$ Bodies are moving in perpendicular direction,

On $x$ -axis, let speed after collision in $x$ -direction is $V_{1}$ , ${mV}_{0} + 0 = 2 {mV}_{1}$
$V_{1} = V_{0} / 2$
On y-axis, let speed after collision in y-direction is $V_{2}$ ,
$m (2 V_{0}) = 2 {mv}_{2}$
$V_{2} = V_{0}$
Hence, net speed after collision
$V_{net} = \sqrt{V_{1}^{2} + V_{2}^{2}}$
$= \sqrt{{(V_{0} / 2)}^{2} + V_{0}^{2}}$
$= \sqrt{\frac{V_{0}^{2} + 4 V_{0}^{2}}{4}}$
$V_{net} = \frac{V_{0} \sqrt{5}}{2}$

$\tan θ = \frac{V_{0}}{V_{0} / 2} = 2$
$θ = {63.43}^{\circ}$

TS EAMCET 31.07.2022

Work, Energy and Power

149167 A body of mass ' $M$ ' collides against a wall with a speed ' $v$ ' and retraces its path with the same speed, the change in momentum is

1 Zero

2

2 Mv

3

Mv

4

- Mv

Explanation:

B Given, mass $= M$ , speed $= v$
Assuming ball was travelling in $(+ X)$ direction initially, Then, initial momentum $P_{i} = Mv$
After collision ball will move in $(- X)$ direction,
Then, final momentum $P_{f} = - Mv$
Now, change in momentum $= P_{i} - P_{f}$
$= Mv - (- Mv)$
$= 2 Mv$

TS EAMCET 31.07.2022

Work, Energy and Power

149164 A ball of mass $0.15 kg$ hits the wall with its initial speed of $12 {ms}^{- 1}$ and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is $100 N$ . calculate the time duration of the contact of ball with the wall.

1

0.018 s

2

0.036 s

3

0.009 s

4

0.072 s

Explanation:

B Given,
$Mass of ball (m) = 0.15 kg$
$Initial speed of ball (u) = 12 m / s$
$v = - 12 m / s, F = 100 N, Δ t = ?$
Initial momentum $(p_{i}) = mu = 12 \times 0.15$
$p_{i} = 1.8 kgm / s$
Final momentum $(p_{f}) = mv = - 12 \times 0.15$
$p_{f} = - 1.8 kgm / s$
Change in momentum,
$Δ p = p_{i} - p_{f} = 1.8 - (- 1.8)$
$Δ p = 3.6 kg m / s$
We know that,
Change in momentum $(Δ p) = F \times Δ t$
$Δ t = \frac{Δ P}{F} = \frac{3.6}{100} = 0.036 s$

JEE Main-26.07.2022

Work, Energy and Power

149165 Two objects I and II of mass $0.8 kg$ and $0.6 kg$ collide elastically. If before collision, the object II was at rest and object I was moving. Find the ratio of the final velocity of object $I$ and object II.

1

6 : 7

2

4 : 3

3

1 : 8

4

1 : 16

Explanation:

C Given, $m_{1} = 0.8 kg, m_{2} = 0.6 kg$ ,
$u_{2} = 0$
$e = 1 (∵ elastically collide)$
Law of conservation of momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$0.8 u_{1} + 0 = 0.8 v_{1} + 0.6 v_{2}$
$0.8 v_{1} + 0.6 v_{2} = 0.8 u_{1}$
Newton's experimental law
$v_{2} - v_{1} = e (u_{1} - u_{2}) (∵ e = 1)$
$v_{2} - v_{1} = 1 (u_{1} - 0)$
$u_{1} = v_{2} - v_{1}$
From equation (i), we get
$0.8 v_{1} + 0.6 v_{2} = 0.8 (v_{2} - v_{1})$
$0.8 v_{1} + 0.6 v_{2} = 0.8 v_{2} - 0.8 v_{1}$
$1.6 v_{1} = 0.2 v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{2}{16}$
$v_{1} : v_{2} = 1 : 8$

TS EAMCET 30.07.2022

Work, Energy and Power

149166 Two identical bodies of same mass move with speed $V_{0}$ an $2 V_{0}$ , respectively in directions perpendicular to each other. After some time, the bodies collide and merge with each other. The speed after collision is

1

\frac{V_{0}}{2}

2

\frac{V_{0} \sqrt{5}}{2}

3

\frac{V_{0}}{2 \sqrt{5}}

4

V_{0}

Explanation:

B Given,
Two identical bodies of same mass i.e. $m_{1} = m_{2} = m$ Moving with speed $V_{0}$ and $2 V_{0}$
$∵$ Bodies are moving in perpendicular direction,

On $x$ -axis, let speed after collision in $x$ -direction is $V_{1}$ , ${mV}_{0} + 0 = 2 {mV}_{1}$
$V_{1} = V_{0} / 2$
On y-axis, let speed after collision in y-direction is $V_{2}$ ,
$m (2 V_{0}) = 2 {mv}_{2}$
$V_{2} = V_{0}$
Hence, net speed after collision
$V_{net} = \sqrt{V_{1}^{2} + V_{2}^{2}}$
$= \sqrt{{(V_{0} / 2)}^{2} + V_{0}^{2}}$
$= \sqrt{\frac{V_{0}^{2} + 4 V_{0}^{2}}{4}}$
$V_{net} = \frac{V_{0} \sqrt{5}}{2}$

$\tan θ = \frac{V_{0}}{V_{0} / 2} = 2$
$θ = {63.43}^{\circ}$

TS EAMCET 31.07.2022

Work, Energy and Power

149167 A body of mass ' $M$ ' collides against a wall with a speed ' $v$ ' and retraces its path with the same speed, the change in momentum is

1 Zero

2

2 Mv

3

Mv

4

- Mv

Explanation:

B Given, mass $= M$ , speed $= v$
Assuming ball was travelling in $(+ X)$ direction initially, Then, initial momentum $P_{i} = Mv$
After collision ball will move in $(- X)$ direction,
Then, final momentum $P_{f} = - Mv$
Now, change in momentum $= P_{i} - P_{f}$
$= Mv - (- Mv)$
$= 2 Mv$

TS EAMCET 31.07.2022

Work, Energy and Power

149164 A ball of mass $0.15 kg$ hits the wall with its initial speed of $12 {ms}^{- 1}$ and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is $100 N$ . calculate the time duration of the contact of ball with the wall.

1

0.018 s

2

0.036 s

3

0.009 s

4

0.072 s

Explanation:

B Given,
$Mass of ball (m) = 0.15 kg$
$Initial speed of ball (u) = 12 m / s$
$v = - 12 m / s, F = 100 N, Δ t = ?$
Initial momentum $(p_{i}) = mu = 12 \times 0.15$
$p_{i} = 1.8 kgm / s$
Final momentum $(p_{f}) = mv = - 12 \times 0.15$
$p_{f} = - 1.8 kgm / s$
Change in momentum,
$Δ p = p_{i} - p_{f} = 1.8 - (- 1.8)$
$Δ p = 3.6 kg m / s$
We know that,
Change in momentum $(Δ p) = F \times Δ t$
$Δ t = \frac{Δ P}{F} = \frac{3.6}{100} = 0.036 s$

JEE Main-26.07.2022

Work, Energy and Power

149165 Two objects I and II of mass $0.8 kg$ and $0.6 kg$ collide elastically. If before collision, the object II was at rest and object I was moving. Find the ratio of the final velocity of object $I$ and object II.

1

6 : 7

2

4 : 3

3

1 : 8

4

1 : 16

Explanation:

C Given, $m_{1} = 0.8 kg, m_{2} = 0.6 kg$ ,
$u_{2} = 0$
$e = 1 (∵ elastically collide)$
Law of conservation of momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$0.8 u_{1} + 0 = 0.8 v_{1} + 0.6 v_{2}$
$0.8 v_{1} + 0.6 v_{2} = 0.8 u_{1}$
Newton's experimental law
$v_{2} - v_{1} = e (u_{1} - u_{2}) (∵ e = 1)$
$v_{2} - v_{1} = 1 (u_{1} - 0)$
$u_{1} = v_{2} - v_{1}$
From equation (i), we get
$0.8 v_{1} + 0.6 v_{2} = 0.8 (v_{2} - v_{1})$
$0.8 v_{1} + 0.6 v_{2} = 0.8 v_{2} - 0.8 v_{1}$
$1.6 v_{1} = 0.2 v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{2}{16}$
$v_{1} : v_{2} = 1 : 8$

TS EAMCET 30.07.2022

Work, Energy and Power

149166 Two identical bodies of same mass move with speed $V_{0}$ an $2 V_{0}$ , respectively in directions perpendicular to each other. After some time, the bodies collide and merge with each other. The speed after collision is

1

\frac{V_{0}}{2}

2

\frac{V_{0} \sqrt{5}}{2}

3

\frac{V_{0}}{2 \sqrt{5}}

4

V_{0}

Explanation:

B Given,
Two identical bodies of same mass i.e. $m_{1} = m_{2} = m$ Moving with speed $V_{0}$ and $2 V_{0}$
$∵$ Bodies are moving in perpendicular direction,

On $x$ -axis, let speed after collision in $x$ -direction is $V_{1}$ , ${mV}_{0} + 0 = 2 {mV}_{1}$
$V_{1} = V_{0} / 2$
On y-axis, let speed after collision in y-direction is $V_{2}$ ,
$m (2 V_{0}) = 2 {mv}_{2}$
$V_{2} = V_{0}$
Hence, net speed after collision
$V_{net} = \sqrt{V_{1}^{2} + V_{2}^{2}}$
$= \sqrt{{(V_{0} / 2)}^{2} + V_{0}^{2}}$
$= \sqrt{\frac{V_{0}^{2} + 4 V_{0}^{2}}{4}}$
$V_{net} = \frac{V_{0} \sqrt{5}}{2}$

$\tan θ = \frac{V_{0}}{V_{0} / 2} = 2$
$θ = {63.43}^{\circ}$

TS EAMCET 31.07.2022

Work, Energy and Power

149167 A body of mass ' $M$ ' collides against a wall with a speed ' $v$ ' and retraces its path with the same speed, the change in momentum is

1 Zero

2

2 Mv

3

Mv

4

- Mv

Explanation:

B Given, mass $= M$ , speed $= v$
Assuming ball was travelling in $(+ X)$ direction initially, Then, initial momentum $P_{i} = Mv$
After collision ball will move in $(- X)$ direction,
Then, final momentum $P_{f} = - Mv$
Now, change in momentum $= P_{i} - P_{f}$
$= Mv - (- Mv)$
$= 2 Mv$

TS EAMCET 31.07.2022

Work, Energy and Power

149164 A ball of mass $0.15 kg$ hits the wall with its initial speed of $12 {ms}^{- 1}$ and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is $100 N$ . calculate the time duration of the contact of ball with the wall.

1

0.018 s

2

0.036 s

3

0.009 s

4

0.072 s

Explanation:

B Given,
$Mass of ball (m) = 0.15 kg$
$Initial speed of ball (u) = 12 m / s$
$v = - 12 m / s, F = 100 N, Δ t = ?$
Initial momentum $(p_{i}) = mu = 12 \times 0.15$
$p_{i} = 1.8 kgm / s$
Final momentum $(p_{f}) = mv = - 12 \times 0.15$
$p_{f} = - 1.8 kgm / s$
Change in momentum,
$Δ p = p_{i} - p_{f} = 1.8 - (- 1.8)$
$Δ p = 3.6 kg m / s$
We know that,
Change in momentum $(Δ p) = F \times Δ t$
$Δ t = \frac{Δ P}{F} = \frac{3.6}{100} = 0.036 s$

JEE Main-26.07.2022

Work, Energy and Power

149165 Two objects I and II of mass $0.8 kg$ and $0.6 kg$ collide elastically. If before collision, the object II was at rest and object I was moving. Find the ratio of the final velocity of object $I$ and object II.

1

6 : 7

2

4 : 3

3

1 : 8

4

1 : 16

Explanation:

C Given, $m_{1} = 0.8 kg, m_{2} = 0.6 kg$ ,
$u_{2} = 0$
$e = 1 (∵ elastically collide)$
Law of conservation of momentum
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$0.8 u_{1} + 0 = 0.8 v_{1} + 0.6 v_{2}$
$0.8 v_{1} + 0.6 v_{2} = 0.8 u_{1}$
Newton's experimental law
$v_{2} - v_{1} = e (u_{1} - u_{2}) (∵ e = 1)$
$v_{2} - v_{1} = 1 (u_{1} - 0)$
$u_{1} = v_{2} - v_{1}$
From equation (i), we get
$0.8 v_{1} + 0.6 v_{2} = 0.8 (v_{2} - v_{1})$
$0.8 v_{1} + 0.6 v_{2} = 0.8 v_{2} - 0.8 v_{1}$
$1.6 v_{1} = 0.2 v_{2}$
$\frac{v_{1}}{v_{2}} = \frac{2}{16}$
$v_{1} : v_{2} = 1 : 8$

TS EAMCET 30.07.2022

Work, Energy and Power

149166 Two identical bodies of same mass move with speed $V_{0}$ an $2 V_{0}$ , respectively in directions perpendicular to each other. After some time, the bodies collide and merge with each other. The speed after collision is

1

\frac{V_{0}}{2}

2

\frac{V_{0} \sqrt{5}}{2}

3

\frac{V_{0}}{2 \sqrt{5}}

4

V_{0}

Explanation:

B Given,
Two identical bodies of same mass i.e. $m_{1} = m_{2} = m$ Moving with speed $V_{0}$ and $2 V_{0}$
$∵$ Bodies are moving in perpendicular direction,

On $x$ -axis, let speed after collision in $x$ -direction is $V_{1}$ , ${mV}_{0} + 0 = 2 {mV}_{1}$
$V_{1} = V_{0} / 2$
On y-axis, let speed after collision in y-direction is $V_{2}$ ,
$m (2 V_{0}) = 2 {mv}_{2}$
$V_{2} = V_{0}$
Hence, net speed after collision
$V_{net} = \sqrt{V_{1}^{2} + V_{2}^{2}}$
$= \sqrt{{(V_{0} / 2)}^{2} + V_{0}^{2}}$
$= \sqrt{\frac{V_{0}^{2} + 4 V_{0}^{2}}{4}}$
$V_{net} = \frac{V_{0} \sqrt{5}}{2}$

$\tan θ = \frac{V_{0}}{V_{0} / 2} = 2$
$θ = {63.43}^{\circ}$

TS EAMCET 31.07.2022

Work, Energy and Power

149167 A body of mass ' $M$ ' collides against a wall with a speed ' $v$ ' and retraces its path with the same speed, the change in momentum is

1 Zero

2

2 Mv

3

Mv

4

- Mv

Explanation:

B Given, mass $= M$ , speed $= v$
Assuming ball was travelling in $(+ X)$ direction initially, Then, initial momentum $P_{i} = Mv$
After collision ball will move in $(- X)$ direction,
Then, final momentum $P_{f} = - Mv$
Now, change in momentum $= P_{i} - P_{f}$
$= Mv - (- Mv)$
$= 2 Mv$

TS EAMCET 31.07.2022

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