QBManager

Work, Energy and Power

148912 Two masses of 1 gram and 4 gram are moving with equal kinetic energy. The ratio of the magnitudes of their momenta is

1

1 : 16

2

1 : 2

3

2 : 1

4

4 : 1

Explanation:

B Given, $m_{1} = 1 gm, m_{2} = 4 gm$
We know that, K.E. $= \frac{1}{2} {mv}^{2}$
$K.E. = \frac{1}{2} m \cdot \frac{p^{2}}{m^{2}} (∵ p = mv \Rightarrow v = \frac{p}{m})$
$= \frac{1}{2} \frac{p^{2}}{m}$
According to question both mass have same kinetic energy,
$So, \frac{1}{2} \frac{p_{1}^{2}}{m_{1}} = \frac{1}{2} \frac{p_{2}^{2}}{m_{2}}$
$\frac{p_{1}^{2}}{1} = \frac{p_{2}^{2}}{4}$
$∴ \frac{p_{1}}{p_{2}} = \frac{1}{2}$

MHT-CET 2020

Work, Energy and Power

148913 If we increase kinetic energy of a body $300 %$ , then percent increase in its momentum is

1

50 %

2

300 %

3

100 %

4

150 %

Explanation:

C We know that,
Momentum $(p) = \sqrt{2 mK} (K =$ kinetic energy $)$
$K = \frac{p^{2}}{2 m}$
If kinetic energy of a body is increased by $300 %$ , let its momentum becomes $p$ '.
$K^{'} = K + \frac{300}{100} K = 4 K [\begin{array}{l} Initial K.E. = K \\ Final K.E. = K^{'} \end{array}]$
Therefore, momentum is given by.
$p^{'} = \sqrt{2 m \times K^{'}}$
$p^{'} = \sqrt{2 m \times 4 K} [\begin{array}{l} Initial momentum = p \\ Finalmomentum = p^{'} \end{array}]$
$= 2 \sqrt{2 mK}$
$p^{'} = 2 p$
$∴ Percentage change in momentum.$
$\frac{Δ p}{p} \times 100 = \frac{p^{'} - p}{p} \times 100$
$= (\frac{p^{'}}{p} - 1) \times 100$
$= (\frac{2 p}{p} - 1) \times 100$
$= 100 %$
Hence, momentum will increase by $100 %$ .

JIPMER-2014

Work, Energy and Power

148914 If the momentum of a body moving in a straight line is increased by $50 %$ , the percent increase in its kinetic energy will be

1

75 %

2

100 %

3

125 %

4

150 %

Explanation:

C Kinetic energy $(KE)_{1} = \frac{p^{2}}{2 m}$
When momentum is increased by $50 %$ then $p^{'} = 1.5 p$
$∴$ Kinetic energy $(KE)_{2} = \frac{(1.5 p)^{2}}{2 m} = \frac{2.25 p^{2}}{2 m}$
Increase in kinetic energy $= (KE)_{2} - (KE)_{1}$
$= \frac{2.25 p^{2}}{2 m} - \frac{p^{2}}{2 m} = \frac{1.25 p^{2}}{2 m}$
$∴ %$ increase in kinetic energy $= \frac{(KE)_{2} - (KE)_{1}}{(KE)_{1}} \times 100$
$= \frac{\frac{1.25 p^{2}}{2 m}}{\frac{p^{2}}{2 m}} \times 100$

JEE Main 29.07.2022 Shift-II]

Work, Energy and Power

148915 If the linear momentum of a body is increased by $50 %$ , then the kinetic energy of that body increases by :

1

100 %

2

125 %

3

225 %

4

25 %

Explanation:

B Relation between kinetic energy and momentum
$K.E. = \frac{p^{2}}{2 m}$
According to the question momentum is increased by $50 %$
So, new momentum becomes
$p^{'} = p + 50 % of p$
$p^{'} = p + p \times \frac{50}{100}$
$p^{'} = 1.5 p$
Therefore new kinetic energy,
$K.E.^{'} = \frac{(1.5 p)^{2}}{2 m}$
$K.E.' = \frac{2.25 p^{2}}{2 m}$
Change in kinetic energy becomes.
$= K.E.'-K.E.$
$= \frac{2.25 p^{2}}{2 m} - \frac{p^{2}}{2 m}$
$= \frac{p^{2}}{2 m} (2.25 - 1)$
$= \frac{p^{2}}{2 m} (1.25)$
$= 1.25 K . E . [∵ From equation (i)]$
$Increased in percentage = \frac{Δ K . E .}{K . E .} \times 100 = \frac{1.25 K . E .}{K . E .} \times 100$
$= 125 %$

TS EAMCET 29.09.2020 Shift - I]

Work, Energy and Power

148916 A cricket ball is hit at an angle of $30^{\circ}$ to the horizontal with a kinetic energy $E$ . Its kinetic energy when it reaches the highest point is

1

\frac{E}{2}

2 0

3

\frac{2 E}{3}

4

\frac{3 E}{4}

5

E

Explanation:

D Given, $θ = 30^{\circ}$
We know that, kinetic energy $($ K.E. $=$ E)
$E = \frac{1}{2} {mv}^{2}$
At the highest point vertical component of velocity is zero.
Hence, $u_{y} = 0$ and $u_{x} = v \cos θ$
Then, kinetic energy at highest point $(E_{H})$
$E_{H} = \frac{1}{2} m (v \cos θ)^{2} [From equation (i)]$
$E_{H} = \frac{1}{2} m \times {(v \cos 30^{\circ})}^{2}$
$E_{H} = \frac{1}{2} {mv}^{2} \times {(\frac{\sqrt{3}}{2})}^{2}$
$E_{H} = \frac{3 E}{4}$

TS EAMCET - 09.09.2020 Shift-I]

Work, Energy and Power

148912 Two masses of 1 gram and 4 gram are moving with equal kinetic energy. The ratio of the magnitudes of their momenta is

1

1 : 16

2

1 : 2

3

2 : 1

4

4 : 1

Explanation:

B Given, $m_{1} = 1 gm, m_{2} = 4 gm$
We know that, K.E. $= \frac{1}{2} {mv}^{2}$
$K.E. = \frac{1}{2} m \cdot \frac{p^{2}}{m^{2}} (∵ p = mv \Rightarrow v = \frac{p}{m})$
$= \frac{1}{2} \frac{p^{2}}{m}$
According to question both mass have same kinetic energy,
$So, \frac{1}{2} \frac{p_{1}^{2}}{m_{1}} = \frac{1}{2} \frac{p_{2}^{2}}{m_{2}}$
$\frac{p_{1}^{2}}{1} = \frac{p_{2}^{2}}{4}$
$∴ \frac{p_{1}}{p_{2}} = \frac{1}{2}$

MHT-CET 2020

Work, Energy and Power

148913 If we increase kinetic energy of a body $300 %$ , then percent increase in its momentum is

1

50 %

2

300 %

3

100 %

4

150 %

Explanation:

C We know that,
Momentum $(p) = \sqrt{2 mK} (K =$ kinetic energy $)$
$K = \frac{p^{2}}{2 m}$
If kinetic energy of a body is increased by $300 %$ , let its momentum becomes $p$ '.
$K^{'} = K + \frac{300}{100} K = 4 K [\begin{array}{l} Initial K.E. = K \\ Final K.E. = K^{'} \end{array}]$
Therefore, momentum is given by.
$p^{'} = \sqrt{2 m \times K^{'}}$
$p^{'} = \sqrt{2 m \times 4 K} [\begin{array}{l} Initial momentum = p \\ Finalmomentum = p^{'} \end{array}]$
$= 2 \sqrt{2 mK}$
$p^{'} = 2 p$
$∴ Percentage change in momentum.$
$\frac{Δ p}{p} \times 100 = \frac{p^{'} - p}{p} \times 100$
$= (\frac{p^{'}}{p} - 1) \times 100$
$= (\frac{2 p}{p} - 1) \times 100$
$= 100 %$
Hence, momentum will increase by $100 %$ .

JIPMER-2014

Work, Energy and Power

148914 If the momentum of a body moving in a straight line is increased by $50 %$ , the percent increase in its kinetic energy will be

1

75 %

2

100 %

3

125 %

4

150 %

Explanation:

C Kinetic energy $(KE)_{1} = \frac{p^{2}}{2 m}$
When momentum is increased by $50 %$ then $p^{'} = 1.5 p$
$∴$ Kinetic energy $(KE)_{2} = \frac{(1.5 p)^{2}}{2 m} = \frac{2.25 p^{2}}{2 m}$
Increase in kinetic energy $= (KE)_{2} - (KE)_{1}$
$= \frac{2.25 p^{2}}{2 m} - \frac{p^{2}}{2 m} = \frac{1.25 p^{2}}{2 m}$
$∴ %$ increase in kinetic energy $= \frac{(KE)_{2} - (KE)_{1}}{(KE)_{1}} \times 100$
$= \frac{\frac{1.25 p^{2}}{2 m}}{\frac{p^{2}}{2 m}} \times 100$

JEE Main 29.07.2022 Shift-II]

Work, Energy and Power

148915 If the linear momentum of a body is increased by $50 %$ , then the kinetic energy of that body increases by :

1

100 %

2

125 %

3

225 %

4

25 %

Explanation:

B Relation between kinetic energy and momentum
$K.E. = \frac{p^{2}}{2 m}$
According to the question momentum is increased by $50 %$
So, new momentum becomes
$p^{'} = p + 50 % of p$
$p^{'} = p + p \times \frac{50}{100}$
$p^{'} = 1.5 p$
Therefore new kinetic energy,
$K.E.^{'} = \frac{(1.5 p)^{2}}{2 m}$
$K.E.' = \frac{2.25 p^{2}}{2 m}$
Change in kinetic energy becomes.
$= K.E.'-K.E.$
$= \frac{2.25 p^{2}}{2 m} - \frac{p^{2}}{2 m}$
$= \frac{p^{2}}{2 m} (2.25 - 1)$
$= \frac{p^{2}}{2 m} (1.25)$
$= 1.25 K . E . [∵ From equation (i)]$
$Increased in percentage = \frac{Δ K . E .}{K . E .} \times 100 = \frac{1.25 K . E .}{K . E .} \times 100$
$= 125 %$

TS EAMCET 29.09.2020 Shift - I]

Work, Energy and Power

148916 A cricket ball is hit at an angle of $30^{\circ}$ to the horizontal with a kinetic energy $E$ . Its kinetic energy when it reaches the highest point is

1

\frac{E}{2}

2 0

3

\frac{2 E}{3}

4

\frac{3 E}{4}

5

E

Explanation:

D Given, $θ = 30^{\circ}$
We know that, kinetic energy $($ K.E. $=$ E)
$E = \frac{1}{2} {mv}^{2}$
At the highest point vertical component of velocity is zero.
Hence, $u_{y} = 0$ and $u_{x} = v \cos θ$
Then, kinetic energy at highest point $(E_{H})$
$E_{H} = \frac{1}{2} m (v \cos θ)^{2} [From equation (i)]$
$E_{H} = \frac{1}{2} m \times {(v \cos 30^{\circ})}^{2}$
$E_{H} = \frac{1}{2} {mv}^{2} \times {(\frac{\sqrt{3}}{2})}^{2}$
$E_{H} = \frac{3 E}{4}$

TS EAMCET - 09.09.2020 Shift-I]

Work, Energy and Power

148912 Two masses of 1 gram and 4 gram are moving with equal kinetic energy. The ratio of the magnitudes of their momenta is

1

1 : 16

2

1 : 2

3

2 : 1

4

4 : 1

Explanation:

B Given, $m_{1} = 1 gm, m_{2} = 4 gm$
We know that, K.E. $= \frac{1}{2} {mv}^{2}$
$K.E. = \frac{1}{2} m \cdot \frac{p^{2}}{m^{2}} (∵ p = mv \Rightarrow v = \frac{p}{m})$
$= \frac{1}{2} \frac{p^{2}}{m}$
According to question both mass have same kinetic energy,
$So, \frac{1}{2} \frac{p_{1}^{2}}{m_{1}} = \frac{1}{2} \frac{p_{2}^{2}}{m_{2}}$
$\frac{p_{1}^{2}}{1} = \frac{p_{2}^{2}}{4}$
$∴ \frac{p_{1}}{p_{2}} = \frac{1}{2}$

MHT-CET 2020

Work, Energy and Power

148913 If we increase kinetic energy of a body $300 %$ , then percent increase in its momentum is

1

50 %

2

300 %

3

100 %

4

150 %

Explanation:

C We know that,
Momentum $(p) = \sqrt{2 mK} (K =$ kinetic energy $)$
$K = \frac{p^{2}}{2 m}$
If kinetic energy of a body is increased by $300 %$ , let its momentum becomes $p$ '.
$K^{'} = K + \frac{300}{100} K = 4 K [\begin{array}{l} Initial K.E. = K \\ Final K.E. = K^{'} \end{array}]$
Therefore, momentum is given by.
$p^{'} = \sqrt{2 m \times K^{'}}$
$p^{'} = \sqrt{2 m \times 4 K} [\begin{array}{l} Initial momentum = p \\ Finalmomentum = p^{'} \end{array}]$
$= 2 \sqrt{2 mK}$
$p^{'} = 2 p$
$∴ Percentage change in momentum.$
$\frac{Δ p}{p} \times 100 = \frac{p^{'} - p}{p} \times 100$
$= (\frac{p^{'}}{p} - 1) \times 100$
$= (\frac{2 p}{p} - 1) \times 100$
$= 100 %$
Hence, momentum will increase by $100 %$ .

JIPMER-2014

Work, Energy and Power

148914 If the momentum of a body moving in a straight line is increased by $50 %$ , the percent increase in its kinetic energy will be

1

75 %

2

100 %

3

125 %

4

150 %

Explanation:

C Kinetic energy $(KE)_{1} = \frac{p^{2}}{2 m}$
When momentum is increased by $50 %$ then $p^{'} = 1.5 p$
$∴$ Kinetic energy $(KE)_{2} = \frac{(1.5 p)^{2}}{2 m} = \frac{2.25 p^{2}}{2 m}$
Increase in kinetic energy $= (KE)_{2} - (KE)_{1}$
$= \frac{2.25 p^{2}}{2 m} - \frac{p^{2}}{2 m} = \frac{1.25 p^{2}}{2 m}$
$∴ %$ increase in kinetic energy $= \frac{(KE)_{2} - (KE)_{1}}{(KE)_{1}} \times 100$
$= \frac{\frac{1.25 p^{2}}{2 m}}{\frac{p^{2}}{2 m}} \times 100$

JEE Main 29.07.2022 Shift-II]

Work, Energy and Power

148915 If the linear momentum of a body is increased by $50 %$ , then the kinetic energy of that body increases by :

1

100 %

2

125 %

3

225 %

4

25 %

Explanation:

B Relation between kinetic energy and momentum
$K.E. = \frac{p^{2}}{2 m}$
According to the question momentum is increased by $50 %$
So, new momentum becomes
$p^{'} = p + 50 % of p$
$p^{'} = p + p \times \frac{50}{100}$
$p^{'} = 1.5 p$
Therefore new kinetic energy,
$K.E.^{'} = \frac{(1.5 p)^{2}}{2 m}$
$K.E.' = \frac{2.25 p^{2}}{2 m}$
Change in kinetic energy becomes.
$= K.E.'-K.E.$
$= \frac{2.25 p^{2}}{2 m} - \frac{p^{2}}{2 m}$
$= \frac{p^{2}}{2 m} (2.25 - 1)$
$= \frac{p^{2}}{2 m} (1.25)$
$= 1.25 K . E . [∵ From equation (i)]$
$Increased in percentage = \frac{Δ K . E .}{K . E .} \times 100 = \frac{1.25 K . E .}{K . E .} \times 100$
$= 125 %$

TS EAMCET 29.09.2020 Shift - I]

Work, Energy and Power

148916 A cricket ball is hit at an angle of $30^{\circ}$ to the horizontal with a kinetic energy $E$ . Its kinetic energy when it reaches the highest point is

1

\frac{E}{2}

2 0

3

\frac{2 E}{3}

4

\frac{3 E}{4}

5

E

Explanation:

D Given, $θ = 30^{\circ}$
We know that, kinetic energy $($ K.E. $=$ E)
$E = \frac{1}{2} {mv}^{2}$
At the highest point vertical component of velocity is zero.
Hence, $u_{y} = 0$ and $u_{x} = v \cos θ$
Then, kinetic energy at highest point $(E_{H})$
$E_{H} = \frac{1}{2} m (v \cos θ)^{2} [From equation (i)]$
$E_{H} = \frac{1}{2} m \times {(v \cos 30^{\circ})}^{2}$
$E_{H} = \frac{1}{2} {mv}^{2} \times {(\frac{\sqrt{3}}{2})}^{2}$
$E_{H} = \frac{3 E}{4}$

TS EAMCET - 09.09.2020 Shift-I]

Work, Energy and Power

148912 Two masses of 1 gram and 4 gram are moving with equal kinetic energy. The ratio of the magnitudes of their momenta is

1

1 : 16

2

1 : 2

3

2 : 1

4

4 : 1

Explanation:

B Given, $m_{1} = 1 gm, m_{2} = 4 gm$
We know that, K.E. $= \frac{1}{2} {mv}^{2}$
$K.E. = \frac{1}{2} m \cdot \frac{p^{2}}{m^{2}} (∵ p = mv \Rightarrow v = \frac{p}{m})$
$= \frac{1}{2} \frac{p^{2}}{m}$
According to question both mass have same kinetic energy,
$So, \frac{1}{2} \frac{p_{1}^{2}}{m_{1}} = \frac{1}{2} \frac{p_{2}^{2}}{m_{2}}$
$\frac{p_{1}^{2}}{1} = \frac{p_{2}^{2}}{4}$
$∴ \frac{p_{1}}{p_{2}} = \frac{1}{2}$

MHT-CET 2020

Work, Energy and Power

148913 If we increase kinetic energy of a body $300 %$ , then percent increase in its momentum is

1

50 %

2

300 %

3

100 %

4

150 %

Explanation:

C We know that,
Momentum $(p) = \sqrt{2 mK} (K =$ kinetic energy $)$
$K = \frac{p^{2}}{2 m}$
If kinetic energy of a body is increased by $300 %$ , let its momentum becomes $p$ '.
$K^{'} = K + \frac{300}{100} K = 4 K [\begin{array}{l} Initial K.E. = K \\ Final K.E. = K^{'} \end{array}]$
Therefore, momentum is given by.
$p^{'} = \sqrt{2 m \times K^{'}}$
$p^{'} = \sqrt{2 m \times 4 K} [\begin{array}{l} Initial momentum = p \\ Finalmomentum = p^{'} \end{array}]$
$= 2 \sqrt{2 mK}$
$p^{'} = 2 p$
$∴ Percentage change in momentum.$
$\frac{Δ p}{p} \times 100 = \frac{p^{'} - p}{p} \times 100$
$= (\frac{p^{'}}{p} - 1) \times 100$
$= (\frac{2 p}{p} - 1) \times 100$
$= 100 %$
Hence, momentum will increase by $100 %$ .

JIPMER-2014

Work, Energy and Power

148914 If the momentum of a body moving in a straight line is increased by $50 %$ , the percent increase in its kinetic energy will be

1

75 %

2

100 %

3

125 %

4

150 %

Explanation:

C Kinetic energy $(KE)_{1} = \frac{p^{2}}{2 m}$
When momentum is increased by $50 %$ then $p^{'} = 1.5 p$
$∴$ Kinetic energy $(KE)_{2} = \frac{(1.5 p)^{2}}{2 m} = \frac{2.25 p^{2}}{2 m}$
Increase in kinetic energy $= (KE)_{2} - (KE)_{1}$
$= \frac{2.25 p^{2}}{2 m} - \frac{p^{2}}{2 m} = \frac{1.25 p^{2}}{2 m}$
$∴ %$ increase in kinetic energy $= \frac{(KE)_{2} - (KE)_{1}}{(KE)_{1}} \times 100$
$= \frac{\frac{1.25 p^{2}}{2 m}}{\frac{p^{2}}{2 m}} \times 100$

JEE Main 29.07.2022 Shift-II]

Work, Energy and Power

148915 If the linear momentum of a body is increased by $50 %$ , then the kinetic energy of that body increases by :

1

100 %

2

125 %

3

225 %

4

25 %

Explanation:

B Relation between kinetic energy and momentum
$K.E. = \frac{p^{2}}{2 m}$
According to the question momentum is increased by $50 %$
So, new momentum becomes
$p^{'} = p + 50 % of p$
$p^{'} = p + p \times \frac{50}{100}$
$p^{'} = 1.5 p$
Therefore new kinetic energy,
$K.E.^{'} = \frac{(1.5 p)^{2}}{2 m}$
$K.E.' = \frac{2.25 p^{2}}{2 m}$
Change in kinetic energy becomes.
$= K.E.'-K.E.$
$= \frac{2.25 p^{2}}{2 m} - \frac{p^{2}}{2 m}$
$= \frac{p^{2}}{2 m} (2.25 - 1)$
$= \frac{p^{2}}{2 m} (1.25)$
$= 1.25 K . E . [∵ From equation (i)]$
$Increased in percentage = \frac{Δ K . E .}{K . E .} \times 100 = \frac{1.25 K . E .}{K . E .} \times 100$
$= 125 %$

TS EAMCET 29.09.2020 Shift - I]

Work, Energy and Power

148916 A cricket ball is hit at an angle of $30^{\circ}$ to the horizontal with a kinetic energy $E$ . Its kinetic energy when it reaches the highest point is

1

\frac{E}{2}

2 0

3

\frac{2 E}{3}

4

\frac{3 E}{4}

5

E

Explanation:

D Given, $θ = 30^{\circ}$
We know that, kinetic energy $($ K.E. $=$ E)
$E = \frac{1}{2} {mv}^{2}$
At the highest point vertical component of velocity is zero.
Hence, $u_{y} = 0$ and $u_{x} = v \cos θ$
Then, kinetic energy at highest point $(E_{H})$
$E_{H} = \frac{1}{2} m (v \cos θ)^{2} [From equation (i)]$
$E_{H} = \frac{1}{2} m \times {(v \cos 30^{\circ})}^{2}$
$E_{H} = \frac{1}{2} {mv}^{2} \times {(\frac{\sqrt{3}}{2})}^{2}$
$E_{H} = \frac{3 E}{4}$

TS EAMCET - 09.09.2020 Shift-I]

Work, Energy and Power

148912 Two masses of 1 gram and 4 gram are moving with equal kinetic energy. The ratio of the magnitudes of their momenta is

1

1 : 16

2

1 : 2

3

2 : 1

4

4 : 1

Explanation:

B Given, $m_{1} = 1 gm, m_{2} = 4 gm$
We know that, K.E. $= \frac{1}{2} {mv}^{2}$
$K.E. = \frac{1}{2} m \cdot \frac{p^{2}}{m^{2}} (∵ p = mv \Rightarrow v = \frac{p}{m})$
$= \frac{1}{2} \frac{p^{2}}{m}$
According to question both mass have same kinetic energy,
$So, \frac{1}{2} \frac{p_{1}^{2}}{m_{1}} = \frac{1}{2} \frac{p_{2}^{2}}{m_{2}}$
$\frac{p_{1}^{2}}{1} = \frac{p_{2}^{2}}{4}$
$∴ \frac{p_{1}}{p_{2}} = \frac{1}{2}$

MHT-CET 2020

Work, Energy and Power

148913 If we increase kinetic energy of a body $300 %$ , then percent increase in its momentum is

1

50 %

2

300 %

3

100 %

4

150 %

Explanation:

C We know that,
Momentum $(p) = \sqrt{2 mK} (K =$ kinetic energy $)$
$K = \frac{p^{2}}{2 m}$
If kinetic energy of a body is increased by $300 %$ , let its momentum becomes $p$ '.
$K^{'} = K + \frac{300}{100} K = 4 K [\begin{array}{l} Initial K.E. = K \\ Final K.E. = K^{'} \end{array}]$
Therefore, momentum is given by.
$p^{'} = \sqrt{2 m \times K^{'}}$
$p^{'} = \sqrt{2 m \times 4 K} [\begin{array}{l} Initial momentum = p \\ Finalmomentum = p^{'} \end{array}]$
$= 2 \sqrt{2 mK}$
$p^{'} = 2 p$
$∴ Percentage change in momentum.$
$\frac{Δ p}{p} \times 100 = \frac{p^{'} - p}{p} \times 100$
$= (\frac{p^{'}}{p} - 1) \times 100$
$= (\frac{2 p}{p} - 1) \times 100$
$= 100 %$
Hence, momentum will increase by $100 %$ .

JIPMER-2014

Work, Energy and Power

148914 If the momentum of a body moving in a straight line is increased by $50 %$ , the percent increase in its kinetic energy will be

1

75 %

2

100 %

3

125 %

4

150 %

Explanation:

C Kinetic energy $(KE)_{1} = \frac{p^{2}}{2 m}$
When momentum is increased by $50 %$ then $p^{'} = 1.5 p$
$∴$ Kinetic energy $(KE)_{2} = \frac{(1.5 p)^{2}}{2 m} = \frac{2.25 p^{2}}{2 m}$
Increase in kinetic energy $= (KE)_{2} - (KE)_{1}$
$= \frac{2.25 p^{2}}{2 m} - \frac{p^{2}}{2 m} = \frac{1.25 p^{2}}{2 m}$
$∴ %$ increase in kinetic energy $= \frac{(KE)_{2} - (KE)_{1}}{(KE)_{1}} \times 100$
$= \frac{\frac{1.25 p^{2}}{2 m}}{\frac{p^{2}}{2 m}} \times 100$

JEE Main 29.07.2022 Shift-II]

Work, Energy and Power

148915 If the linear momentum of a body is increased by $50 %$ , then the kinetic energy of that body increases by :

1

100 %

2

125 %

3

225 %

4

25 %

Explanation:

B Relation between kinetic energy and momentum
$K.E. = \frac{p^{2}}{2 m}$
According to the question momentum is increased by $50 %$
So, new momentum becomes
$p^{'} = p + 50 % of p$
$p^{'} = p + p \times \frac{50}{100}$
$p^{'} = 1.5 p$
Therefore new kinetic energy,
$K.E.^{'} = \frac{(1.5 p)^{2}}{2 m}$
$K.E.' = \frac{2.25 p^{2}}{2 m}$
Change in kinetic energy becomes.
$= K.E.'-K.E.$
$= \frac{2.25 p^{2}}{2 m} - \frac{p^{2}}{2 m}$
$= \frac{p^{2}}{2 m} (2.25 - 1)$
$= \frac{p^{2}}{2 m} (1.25)$
$= 1.25 K . E . [∵ From equation (i)]$
$Increased in percentage = \frac{Δ K . E .}{K . E .} \times 100 = \frac{1.25 K . E .}{K . E .} \times 100$
$= 125 %$

TS EAMCET 29.09.2020 Shift - I]

Work, Energy and Power

148916 A cricket ball is hit at an angle of $30^{\circ}$ to the horizontal with a kinetic energy $E$ . Its kinetic energy when it reaches the highest point is

1

\frac{E}{2}

2 0

3

\frac{2 E}{3}

4

\frac{3 E}{4}

5

E

Explanation:

D Given, $θ = 30^{\circ}$
We know that, kinetic energy $($ K.E. $=$ E)
$E = \frac{1}{2} {mv}^{2}$
At the highest point vertical component of velocity is zero.
Hence, $u_{y} = 0$ and $u_{x} = v \cos θ$
Then, kinetic energy at highest point $(E_{H})$
$E_{H} = \frac{1}{2} m (v \cos θ)^{2} [From equation (i)]$
$E_{H} = \frac{1}{2} m \times {(v \cos 30^{\circ})}^{2}$
$E_{H} = \frac{1}{2} {mv}^{2} \times {(\frac{\sqrt{3}}{2})}^{2}$
$E_{H} = \frac{3 E}{4}$

TS EAMCET - 09.09.2020 Shift-I]