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, $\mathrm{m}_{1}=1 \mathrm{gm}, \mathrm{m}_{2}=4 \mathrm{gm}$ We know that, K.E. $=\frac{1}{2} \mathrm{mv}^{2}$ $\text { K.E. } =\frac{1}{2} \mathrm{~m} \cdot \frac{\mathrm{p}^{2}}{\mathrm{~m}^{2}}\left(\because \mathrm{p}=\mathrm{mv} \Rightarrow \mathrm{v}=\frac{\mathrm{p}}{\mathrm{m}}\right)$ $=\frac{1}{2} \frac{\mathrm{p}^{2}}{\mathrm{~m}}$ According to question both mass have same kinetic energy, $\text { So, } \frac{1}{2} \frac{\mathrm{p}_{1}^{2}}{\mathrm{~m}_{1}}=\frac{1}{2} \frac{\mathrm{p}_{2}^{2}}{\mathrm{~m}_{2}}$ $\frac{\mathrm{p}_{1}^{2}}{1}=\frac{\mathrm{p}_{2}^{2}}{4}$ $\therefore \quad \frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=\frac{1}{2}$
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 $(p)=\sqrt{2 \mathrm{mK}}(\mathrm{K}=$ kinetic energy $)$ $\mathrm{K}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ If kinetic energy of a body is increased by $300 \%$, let its momentum becomes $\mathrm{p}$ '. \(\mathrm{K}^{\prime}=\mathrm{K}+\frac{300}{100} \mathrm{~K}=4 \mathrm{~K} \quad\left[\begin{array}{l}\text { Initial K.E. }=\mathrm{K} \\ \text { Final K.E. }=\mathrm{K}^{\prime}\end{array}\right]\) Therefore, momentum is given by. $\mathrm{p}^{\prime}=\sqrt{2 \mathrm{~m} \times \mathrm{K}^{\prime}}$ \(\mathrm{p}^{\prime}=\sqrt{2 \mathrm{~m} \times 4 \mathrm{~K}} \quad\left[\begin{array}{l}\text { Initial momentum }=\mathrm{p} \\ \text { Finalmomentum }=\mathrm{p}^{\prime}\end{array}\right]\) $=2 \sqrt{2 \mathrm{mK}}$ $\mathrm{p}^{\prime}=2 \mathrm{p}$ $\therefore \quad \text { Percentage change in momentum. }$ $\frac{\Delta \mathrm{p}}{\mathrm{p}} \times 100=\frac{\mathrm{p}^{\prime}-\mathrm{p}}{\mathrm{p}} \times 100$ $=\left(\frac{\mathrm{p}^{\prime}}{\mathrm{p}}-1\right) \times 100$ $=\left(\frac{2 \mathrm{p}}{\mathrm{p}}-1\right) \times 100$ $=100 \%$ 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 $(\mathrm{KE})_{1}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ When momentum is increased by $50 \%$ then $\mathrm{p}^{\prime}=1.5 \mathrm{p}$ $\therefore$ Kinetic energy $(\mathrm{KE})_{2}=\frac{(1.5 \mathrm{p})^{2}}{2 \mathrm{~m}}=\frac{2.25 \mathrm{p}^{2}}{2 \mathrm{~m}}$ Increase in kinetic energy $=(\mathrm{KE})_{2}-(\mathrm{KE})_{1}$ $=\frac{2.25 \mathrm{p}^{2}}{2 \mathrm{~m}}-\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}=\frac{1.25 \mathrm{p}^{2}}{2 \mathrm{~m}}$ $\therefore \%$ increase in kinetic energy $=\frac{(\mathrm{KE})_{2}-(\mathrm{KE})_{1}}{(\mathrm{KE})_{1}} \times 100$ $=\frac{\frac{1.25 \mathrm{p}^{2}}{2 \mathrm{~m}}}{\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}} \times 100$
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 $\text { K.E. }=\frac{p^{2}}{2 m}$ According to the question momentum is increased by $50 \%$ So, new momentum becomes $\mathrm{p}^{\prime} =\mathrm{p}+50 \% \text { of } \mathrm{p}$ $\mathrm{p}^{\prime} =\mathrm{p}+\mathrm{p} \times \frac{50}{100}$ $\mathrm{p}^{\prime} =1.5 \mathrm{p}$ Therefore new kinetic energy, $\text { K.E. }{ }^{\prime}=\frac{(1.5 \mathrm{p})^2}{2 \mathrm{~m}}$ $\text { K.E.' }=\frac{2.25 \mathrm{p}^2}{2 \mathrm{~m}}$ Change in kinetic energy becomes. $\quad=\text { K.E.'-K.E. }$ $\quad=\frac{2.25 \mathrm{p}^2}{2 \mathrm{~m}}-\frac{\mathrm{p}^2}{2 \mathrm{~m}}$ $=\frac{\mathrm{p}^2}{2 \mathrm{~m}}(2.25-1)$ $=\frac{\mathrm{p}^2}{2 \mathrm{~m}}(1.25)$ $=1.25 \mathrm{~K} . \mathrm{E} . \quad[\because \text { From equation }(\mathrm{i})]$ $\text { Increased in percentage }=\frac{\Delta \mathrm{K} . \mathrm{E} .}{\mathrm{K} . \mathrm{E} .} \times 100=\frac{1.25 \mathrm{~K} . \mathrm{E} .}{\mathrm{K} . \mathrm{E} .} \times 100$ $=125 \%$
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 $\frac{2 \mathrm{E}}{3}$
4 $\frac{3 \mathrm{E}}{4}$
5 $\mathrm{E}$
Explanation:
D Given, $\theta=30^{\circ}$ We know that, kinetic energy $($ K.E. $=$ E) $\mathrm{E}=\frac{1}{2} \mathrm{mv}^{2}$ At the highest point vertical component of velocity is zero. Hence, $\mathrm{u}_{\mathrm{y}}=0$ and $\mathrm{u}_{\mathrm{x}}=\mathrm{v} \cos \theta$ Then, kinetic energy at highest point $\left(\mathrm{E}_{\mathrm{H}}\right)$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{~m}(\mathrm{v} \cos \theta)^{2} \quad \text { [From equation (i)] }$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{~m} \times\left(\mathrm{v} \cos 30^{\circ}\right)^{2}$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{mv}^{2} \times\left(\frac{\sqrt{3}}{2}\right)^{2}$ $\mathrm{E}_{\mathrm{H}} =\frac{3 \mathrm{E}}{4}$
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, $\mathrm{m}_{1}=1 \mathrm{gm}, \mathrm{m}_{2}=4 \mathrm{gm}$ We know that, K.E. $=\frac{1}{2} \mathrm{mv}^{2}$ $\text { K.E. } =\frac{1}{2} \mathrm{~m} \cdot \frac{\mathrm{p}^{2}}{\mathrm{~m}^{2}}\left(\because \mathrm{p}=\mathrm{mv} \Rightarrow \mathrm{v}=\frac{\mathrm{p}}{\mathrm{m}}\right)$ $=\frac{1}{2} \frac{\mathrm{p}^{2}}{\mathrm{~m}}$ According to question both mass have same kinetic energy, $\text { So, } \frac{1}{2} \frac{\mathrm{p}_{1}^{2}}{\mathrm{~m}_{1}}=\frac{1}{2} \frac{\mathrm{p}_{2}^{2}}{\mathrm{~m}_{2}}$ $\frac{\mathrm{p}_{1}^{2}}{1}=\frac{\mathrm{p}_{2}^{2}}{4}$ $\therefore \quad \frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=\frac{1}{2}$
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 $(p)=\sqrt{2 \mathrm{mK}}(\mathrm{K}=$ kinetic energy $)$ $\mathrm{K}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ If kinetic energy of a body is increased by $300 \%$, let its momentum becomes $\mathrm{p}$ '. \(\mathrm{K}^{\prime}=\mathrm{K}+\frac{300}{100} \mathrm{~K}=4 \mathrm{~K} \quad\left[\begin{array}{l}\text { Initial K.E. }=\mathrm{K} \\ \text { Final K.E. }=\mathrm{K}^{\prime}\end{array}\right]\) Therefore, momentum is given by. $\mathrm{p}^{\prime}=\sqrt{2 \mathrm{~m} \times \mathrm{K}^{\prime}}$ \(\mathrm{p}^{\prime}=\sqrt{2 \mathrm{~m} \times 4 \mathrm{~K}} \quad\left[\begin{array}{l}\text { Initial momentum }=\mathrm{p} \\ \text { Finalmomentum }=\mathrm{p}^{\prime}\end{array}\right]\) $=2 \sqrt{2 \mathrm{mK}}$ $\mathrm{p}^{\prime}=2 \mathrm{p}$ $\therefore \quad \text { Percentage change in momentum. }$ $\frac{\Delta \mathrm{p}}{\mathrm{p}} \times 100=\frac{\mathrm{p}^{\prime}-\mathrm{p}}{\mathrm{p}} \times 100$ $=\left(\frac{\mathrm{p}^{\prime}}{\mathrm{p}}-1\right) \times 100$ $=\left(\frac{2 \mathrm{p}}{\mathrm{p}}-1\right) \times 100$ $=100 \%$ 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 $(\mathrm{KE})_{1}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ When momentum is increased by $50 \%$ then $\mathrm{p}^{\prime}=1.5 \mathrm{p}$ $\therefore$ Kinetic energy $(\mathrm{KE})_{2}=\frac{(1.5 \mathrm{p})^{2}}{2 \mathrm{~m}}=\frac{2.25 \mathrm{p}^{2}}{2 \mathrm{~m}}$ Increase in kinetic energy $=(\mathrm{KE})_{2}-(\mathrm{KE})_{1}$ $=\frac{2.25 \mathrm{p}^{2}}{2 \mathrm{~m}}-\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}=\frac{1.25 \mathrm{p}^{2}}{2 \mathrm{~m}}$ $\therefore \%$ increase in kinetic energy $=\frac{(\mathrm{KE})_{2}-(\mathrm{KE})_{1}}{(\mathrm{KE})_{1}} \times 100$ $=\frac{\frac{1.25 \mathrm{p}^{2}}{2 \mathrm{~m}}}{\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}} \times 100$
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 $\text { K.E. }=\frac{p^{2}}{2 m}$ According to the question momentum is increased by $50 \%$ So, new momentum becomes $\mathrm{p}^{\prime} =\mathrm{p}+50 \% \text { of } \mathrm{p}$ $\mathrm{p}^{\prime} =\mathrm{p}+\mathrm{p} \times \frac{50}{100}$ $\mathrm{p}^{\prime} =1.5 \mathrm{p}$ Therefore new kinetic energy, $\text { K.E. }{ }^{\prime}=\frac{(1.5 \mathrm{p})^2}{2 \mathrm{~m}}$ $\text { K.E.' }=\frac{2.25 \mathrm{p}^2}{2 \mathrm{~m}}$ Change in kinetic energy becomes. $\quad=\text { K.E.'-K.E. }$ $\quad=\frac{2.25 \mathrm{p}^2}{2 \mathrm{~m}}-\frac{\mathrm{p}^2}{2 \mathrm{~m}}$ $=\frac{\mathrm{p}^2}{2 \mathrm{~m}}(2.25-1)$ $=\frac{\mathrm{p}^2}{2 \mathrm{~m}}(1.25)$ $=1.25 \mathrm{~K} . \mathrm{E} . \quad[\because \text { From equation }(\mathrm{i})]$ $\text { Increased in percentage }=\frac{\Delta \mathrm{K} . \mathrm{E} .}{\mathrm{K} . \mathrm{E} .} \times 100=\frac{1.25 \mathrm{~K} . \mathrm{E} .}{\mathrm{K} . \mathrm{E} .} \times 100$ $=125 \%$
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 $\frac{2 \mathrm{E}}{3}$
4 $\frac{3 \mathrm{E}}{4}$
5 $\mathrm{E}$
Explanation:
D Given, $\theta=30^{\circ}$ We know that, kinetic energy $($ K.E. $=$ E) $\mathrm{E}=\frac{1}{2} \mathrm{mv}^{2}$ At the highest point vertical component of velocity is zero. Hence, $\mathrm{u}_{\mathrm{y}}=0$ and $\mathrm{u}_{\mathrm{x}}=\mathrm{v} \cos \theta$ Then, kinetic energy at highest point $\left(\mathrm{E}_{\mathrm{H}}\right)$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{~m}(\mathrm{v} \cos \theta)^{2} \quad \text { [From equation (i)] }$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{~m} \times\left(\mathrm{v} \cos 30^{\circ}\right)^{2}$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{mv}^{2} \times\left(\frac{\sqrt{3}}{2}\right)^{2}$ $\mathrm{E}_{\mathrm{H}} =\frac{3 \mathrm{E}}{4}$
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, $\mathrm{m}_{1}=1 \mathrm{gm}, \mathrm{m}_{2}=4 \mathrm{gm}$ We know that, K.E. $=\frac{1}{2} \mathrm{mv}^{2}$ $\text { K.E. } =\frac{1}{2} \mathrm{~m} \cdot \frac{\mathrm{p}^{2}}{\mathrm{~m}^{2}}\left(\because \mathrm{p}=\mathrm{mv} \Rightarrow \mathrm{v}=\frac{\mathrm{p}}{\mathrm{m}}\right)$ $=\frac{1}{2} \frac{\mathrm{p}^{2}}{\mathrm{~m}}$ According to question both mass have same kinetic energy, $\text { So, } \frac{1}{2} \frac{\mathrm{p}_{1}^{2}}{\mathrm{~m}_{1}}=\frac{1}{2} \frac{\mathrm{p}_{2}^{2}}{\mathrm{~m}_{2}}$ $\frac{\mathrm{p}_{1}^{2}}{1}=\frac{\mathrm{p}_{2}^{2}}{4}$ $\therefore \quad \frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=\frac{1}{2}$
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 $(p)=\sqrt{2 \mathrm{mK}}(\mathrm{K}=$ kinetic energy $)$ $\mathrm{K}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ If kinetic energy of a body is increased by $300 \%$, let its momentum becomes $\mathrm{p}$ '. \(\mathrm{K}^{\prime}=\mathrm{K}+\frac{300}{100} \mathrm{~K}=4 \mathrm{~K} \quad\left[\begin{array}{l}\text { Initial K.E. }=\mathrm{K} \\ \text { Final K.E. }=\mathrm{K}^{\prime}\end{array}\right]\) Therefore, momentum is given by. $\mathrm{p}^{\prime}=\sqrt{2 \mathrm{~m} \times \mathrm{K}^{\prime}}$ \(\mathrm{p}^{\prime}=\sqrt{2 \mathrm{~m} \times 4 \mathrm{~K}} \quad\left[\begin{array}{l}\text { Initial momentum }=\mathrm{p} \\ \text { Finalmomentum }=\mathrm{p}^{\prime}\end{array}\right]\) $=2 \sqrt{2 \mathrm{mK}}$ $\mathrm{p}^{\prime}=2 \mathrm{p}$ $\therefore \quad \text { Percentage change in momentum. }$ $\frac{\Delta \mathrm{p}}{\mathrm{p}} \times 100=\frac{\mathrm{p}^{\prime}-\mathrm{p}}{\mathrm{p}} \times 100$ $=\left(\frac{\mathrm{p}^{\prime}}{\mathrm{p}}-1\right) \times 100$ $=\left(\frac{2 \mathrm{p}}{\mathrm{p}}-1\right) \times 100$ $=100 \%$ 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 $(\mathrm{KE})_{1}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ When momentum is increased by $50 \%$ then $\mathrm{p}^{\prime}=1.5 \mathrm{p}$ $\therefore$ Kinetic energy $(\mathrm{KE})_{2}=\frac{(1.5 \mathrm{p})^{2}}{2 \mathrm{~m}}=\frac{2.25 \mathrm{p}^{2}}{2 \mathrm{~m}}$ Increase in kinetic energy $=(\mathrm{KE})_{2}-(\mathrm{KE})_{1}$ $=\frac{2.25 \mathrm{p}^{2}}{2 \mathrm{~m}}-\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}=\frac{1.25 \mathrm{p}^{2}}{2 \mathrm{~m}}$ $\therefore \%$ increase in kinetic energy $=\frac{(\mathrm{KE})_{2}-(\mathrm{KE})_{1}}{(\mathrm{KE})_{1}} \times 100$ $=\frac{\frac{1.25 \mathrm{p}^{2}}{2 \mathrm{~m}}}{\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}} \times 100$
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 $\text { K.E. }=\frac{p^{2}}{2 m}$ According to the question momentum is increased by $50 \%$ So, new momentum becomes $\mathrm{p}^{\prime} =\mathrm{p}+50 \% \text { of } \mathrm{p}$ $\mathrm{p}^{\prime} =\mathrm{p}+\mathrm{p} \times \frac{50}{100}$ $\mathrm{p}^{\prime} =1.5 \mathrm{p}$ Therefore new kinetic energy, $\text { K.E. }{ }^{\prime}=\frac{(1.5 \mathrm{p})^2}{2 \mathrm{~m}}$ $\text { K.E.' }=\frac{2.25 \mathrm{p}^2}{2 \mathrm{~m}}$ Change in kinetic energy becomes. $\quad=\text { K.E.'-K.E. }$ $\quad=\frac{2.25 \mathrm{p}^2}{2 \mathrm{~m}}-\frac{\mathrm{p}^2}{2 \mathrm{~m}}$ $=\frac{\mathrm{p}^2}{2 \mathrm{~m}}(2.25-1)$ $=\frac{\mathrm{p}^2}{2 \mathrm{~m}}(1.25)$ $=1.25 \mathrm{~K} . \mathrm{E} . \quad[\because \text { From equation }(\mathrm{i})]$ $\text { Increased in percentage }=\frac{\Delta \mathrm{K} . \mathrm{E} .}{\mathrm{K} . \mathrm{E} .} \times 100=\frac{1.25 \mathrm{~K} . \mathrm{E} .}{\mathrm{K} . \mathrm{E} .} \times 100$ $=125 \%$
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 $\frac{2 \mathrm{E}}{3}$
4 $\frac{3 \mathrm{E}}{4}$
5 $\mathrm{E}$
Explanation:
D Given, $\theta=30^{\circ}$ We know that, kinetic energy $($ K.E. $=$ E) $\mathrm{E}=\frac{1}{2} \mathrm{mv}^{2}$ At the highest point vertical component of velocity is zero. Hence, $\mathrm{u}_{\mathrm{y}}=0$ and $\mathrm{u}_{\mathrm{x}}=\mathrm{v} \cos \theta$ Then, kinetic energy at highest point $\left(\mathrm{E}_{\mathrm{H}}\right)$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{~m}(\mathrm{v} \cos \theta)^{2} \quad \text { [From equation (i)] }$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{~m} \times\left(\mathrm{v} \cos 30^{\circ}\right)^{2}$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{mv}^{2} \times\left(\frac{\sqrt{3}}{2}\right)^{2}$ $\mathrm{E}_{\mathrm{H}} =\frac{3 \mathrm{E}}{4}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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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, $\mathrm{m}_{1}=1 \mathrm{gm}, \mathrm{m}_{2}=4 \mathrm{gm}$ We know that, K.E. $=\frac{1}{2} \mathrm{mv}^{2}$ $\text { K.E. } =\frac{1}{2} \mathrm{~m} \cdot \frac{\mathrm{p}^{2}}{\mathrm{~m}^{2}}\left(\because \mathrm{p}=\mathrm{mv} \Rightarrow \mathrm{v}=\frac{\mathrm{p}}{\mathrm{m}}\right)$ $=\frac{1}{2} \frac{\mathrm{p}^{2}}{\mathrm{~m}}$ According to question both mass have same kinetic energy, $\text { So, } \frac{1}{2} \frac{\mathrm{p}_{1}^{2}}{\mathrm{~m}_{1}}=\frac{1}{2} \frac{\mathrm{p}_{2}^{2}}{\mathrm{~m}_{2}}$ $\frac{\mathrm{p}_{1}^{2}}{1}=\frac{\mathrm{p}_{2}^{2}}{4}$ $\therefore \quad \frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=\frac{1}{2}$
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 $(p)=\sqrt{2 \mathrm{mK}}(\mathrm{K}=$ kinetic energy $)$ $\mathrm{K}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ If kinetic energy of a body is increased by $300 \%$, let its momentum becomes $\mathrm{p}$ '. \(\mathrm{K}^{\prime}=\mathrm{K}+\frac{300}{100} \mathrm{~K}=4 \mathrm{~K} \quad\left[\begin{array}{l}\text { Initial K.E. }=\mathrm{K} \\ \text { Final K.E. }=\mathrm{K}^{\prime}\end{array}\right]\) Therefore, momentum is given by. $\mathrm{p}^{\prime}=\sqrt{2 \mathrm{~m} \times \mathrm{K}^{\prime}}$ \(\mathrm{p}^{\prime}=\sqrt{2 \mathrm{~m} \times 4 \mathrm{~K}} \quad\left[\begin{array}{l}\text { Initial momentum }=\mathrm{p} \\ \text { Finalmomentum }=\mathrm{p}^{\prime}\end{array}\right]\) $=2 \sqrt{2 \mathrm{mK}}$ $\mathrm{p}^{\prime}=2 \mathrm{p}$ $\therefore \quad \text { Percentage change in momentum. }$ $\frac{\Delta \mathrm{p}}{\mathrm{p}} \times 100=\frac{\mathrm{p}^{\prime}-\mathrm{p}}{\mathrm{p}} \times 100$ $=\left(\frac{\mathrm{p}^{\prime}}{\mathrm{p}}-1\right) \times 100$ $=\left(\frac{2 \mathrm{p}}{\mathrm{p}}-1\right) \times 100$ $=100 \%$ 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 $(\mathrm{KE})_{1}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ When momentum is increased by $50 \%$ then $\mathrm{p}^{\prime}=1.5 \mathrm{p}$ $\therefore$ Kinetic energy $(\mathrm{KE})_{2}=\frac{(1.5 \mathrm{p})^{2}}{2 \mathrm{~m}}=\frac{2.25 \mathrm{p}^{2}}{2 \mathrm{~m}}$ Increase in kinetic energy $=(\mathrm{KE})_{2}-(\mathrm{KE})_{1}$ $=\frac{2.25 \mathrm{p}^{2}}{2 \mathrm{~m}}-\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}=\frac{1.25 \mathrm{p}^{2}}{2 \mathrm{~m}}$ $\therefore \%$ increase in kinetic energy $=\frac{(\mathrm{KE})_{2}-(\mathrm{KE})_{1}}{(\mathrm{KE})_{1}} \times 100$ $=\frac{\frac{1.25 \mathrm{p}^{2}}{2 \mathrm{~m}}}{\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}} \times 100$
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 $\text { K.E. }=\frac{p^{2}}{2 m}$ According to the question momentum is increased by $50 \%$ So, new momentum becomes $\mathrm{p}^{\prime} =\mathrm{p}+50 \% \text { of } \mathrm{p}$ $\mathrm{p}^{\prime} =\mathrm{p}+\mathrm{p} \times \frac{50}{100}$ $\mathrm{p}^{\prime} =1.5 \mathrm{p}$ Therefore new kinetic energy, $\text { K.E. }{ }^{\prime}=\frac{(1.5 \mathrm{p})^2}{2 \mathrm{~m}}$ $\text { K.E.' }=\frac{2.25 \mathrm{p}^2}{2 \mathrm{~m}}$ Change in kinetic energy becomes. $\quad=\text { K.E.'-K.E. }$ $\quad=\frac{2.25 \mathrm{p}^2}{2 \mathrm{~m}}-\frac{\mathrm{p}^2}{2 \mathrm{~m}}$ $=\frac{\mathrm{p}^2}{2 \mathrm{~m}}(2.25-1)$ $=\frac{\mathrm{p}^2}{2 \mathrm{~m}}(1.25)$ $=1.25 \mathrm{~K} . \mathrm{E} . \quad[\because \text { From equation }(\mathrm{i})]$ $\text { Increased in percentage }=\frac{\Delta \mathrm{K} . \mathrm{E} .}{\mathrm{K} . \mathrm{E} .} \times 100=\frac{1.25 \mathrm{~K} . \mathrm{E} .}{\mathrm{K} . \mathrm{E} .} \times 100$ $=125 \%$
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 $\frac{2 \mathrm{E}}{3}$
4 $\frac{3 \mathrm{E}}{4}$
5 $\mathrm{E}$
Explanation:
D Given, $\theta=30^{\circ}$ We know that, kinetic energy $($ K.E. $=$ E) $\mathrm{E}=\frac{1}{2} \mathrm{mv}^{2}$ At the highest point vertical component of velocity is zero. Hence, $\mathrm{u}_{\mathrm{y}}=0$ and $\mathrm{u}_{\mathrm{x}}=\mathrm{v} \cos \theta$ Then, kinetic energy at highest point $\left(\mathrm{E}_{\mathrm{H}}\right)$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{~m}(\mathrm{v} \cos \theta)^{2} \quad \text { [From equation (i)] }$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{~m} \times\left(\mathrm{v} \cos 30^{\circ}\right)^{2}$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{mv}^{2} \times\left(\frac{\sqrt{3}}{2}\right)^{2}$ $\mathrm{E}_{\mathrm{H}} =\frac{3 \mathrm{E}}{4}$
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, $\mathrm{m}_{1}=1 \mathrm{gm}, \mathrm{m}_{2}=4 \mathrm{gm}$ We know that, K.E. $=\frac{1}{2} \mathrm{mv}^{2}$ $\text { K.E. } =\frac{1}{2} \mathrm{~m} \cdot \frac{\mathrm{p}^{2}}{\mathrm{~m}^{2}}\left(\because \mathrm{p}=\mathrm{mv} \Rightarrow \mathrm{v}=\frac{\mathrm{p}}{\mathrm{m}}\right)$ $=\frac{1}{2} \frac{\mathrm{p}^{2}}{\mathrm{~m}}$ According to question both mass have same kinetic energy, $\text { So, } \frac{1}{2} \frac{\mathrm{p}_{1}^{2}}{\mathrm{~m}_{1}}=\frac{1}{2} \frac{\mathrm{p}_{2}^{2}}{\mathrm{~m}_{2}}$ $\frac{\mathrm{p}_{1}^{2}}{1}=\frac{\mathrm{p}_{2}^{2}}{4}$ $\therefore \quad \frac{\mathrm{p}_{1}}{\mathrm{p}_{2}}=\frac{1}{2}$
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 $(p)=\sqrt{2 \mathrm{mK}}(\mathrm{K}=$ kinetic energy $)$ $\mathrm{K}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ If kinetic energy of a body is increased by $300 \%$, let its momentum becomes $\mathrm{p}$ '. \(\mathrm{K}^{\prime}=\mathrm{K}+\frac{300}{100} \mathrm{~K}=4 \mathrm{~K} \quad\left[\begin{array}{l}\text { Initial K.E. }=\mathrm{K} \\ \text { Final K.E. }=\mathrm{K}^{\prime}\end{array}\right]\) Therefore, momentum is given by. $\mathrm{p}^{\prime}=\sqrt{2 \mathrm{~m} \times \mathrm{K}^{\prime}}$ \(\mathrm{p}^{\prime}=\sqrt{2 \mathrm{~m} \times 4 \mathrm{~K}} \quad\left[\begin{array}{l}\text { Initial momentum }=\mathrm{p} \\ \text { Finalmomentum }=\mathrm{p}^{\prime}\end{array}\right]\) $=2 \sqrt{2 \mathrm{mK}}$ $\mathrm{p}^{\prime}=2 \mathrm{p}$ $\therefore \quad \text { Percentage change in momentum. }$ $\frac{\Delta \mathrm{p}}{\mathrm{p}} \times 100=\frac{\mathrm{p}^{\prime}-\mathrm{p}}{\mathrm{p}} \times 100$ $=\left(\frac{\mathrm{p}^{\prime}}{\mathrm{p}}-1\right) \times 100$ $=\left(\frac{2 \mathrm{p}}{\mathrm{p}}-1\right) \times 100$ $=100 \%$ 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 $(\mathrm{KE})_{1}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ When momentum is increased by $50 \%$ then $\mathrm{p}^{\prime}=1.5 \mathrm{p}$ $\therefore$ Kinetic energy $(\mathrm{KE})_{2}=\frac{(1.5 \mathrm{p})^{2}}{2 \mathrm{~m}}=\frac{2.25 \mathrm{p}^{2}}{2 \mathrm{~m}}$ Increase in kinetic energy $=(\mathrm{KE})_{2}-(\mathrm{KE})_{1}$ $=\frac{2.25 \mathrm{p}^{2}}{2 \mathrm{~m}}-\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}=\frac{1.25 \mathrm{p}^{2}}{2 \mathrm{~m}}$ $\therefore \%$ increase in kinetic energy $=\frac{(\mathrm{KE})_{2}-(\mathrm{KE})_{1}}{(\mathrm{KE})_{1}} \times 100$ $=\frac{\frac{1.25 \mathrm{p}^{2}}{2 \mathrm{~m}}}{\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}} \times 100$
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 $\text { K.E. }=\frac{p^{2}}{2 m}$ According to the question momentum is increased by $50 \%$ So, new momentum becomes $\mathrm{p}^{\prime} =\mathrm{p}+50 \% \text { of } \mathrm{p}$ $\mathrm{p}^{\prime} =\mathrm{p}+\mathrm{p} \times \frac{50}{100}$ $\mathrm{p}^{\prime} =1.5 \mathrm{p}$ Therefore new kinetic energy, $\text { K.E. }{ }^{\prime}=\frac{(1.5 \mathrm{p})^2}{2 \mathrm{~m}}$ $\text { K.E.' }=\frac{2.25 \mathrm{p}^2}{2 \mathrm{~m}}$ Change in kinetic energy becomes. $\quad=\text { K.E.'-K.E. }$ $\quad=\frac{2.25 \mathrm{p}^2}{2 \mathrm{~m}}-\frac{\mathrm{p}^2}{2 \mathrm{~m}}$ $=\frac{\mathrm{p}^2}{2 \mathrm{~m}}(2.25-1)$ $=\frac{\mathrm{p}^2}{2 \mathrm{~m}}(1.25)$ $=1.25 \mathrm{~K} . \mathrm{E} . \quad[\because \text { From equation }(\mathrm{i})]$ $\text { Increased in percentage }=\frac{\Delta \mathrm{K} . \mathrm{E} .}{\mathrm{K} . \mathrm{E} .} \times 100=\frac{1.25 \mathrm{~K} . \mathrm{E} .}{\mathrm{K} . \mathrm{E} .} \times 100$ $=125 \%$
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 $\frac{2 \mathrm{E}}{3}$
4 $\frac{3 \mathrm{E}}{4}$
5 $\mathrm{E}$
Explanation:
D Given, $\theta=30^{\circ}$ We know that, kinetic energy $($ K.E. $=$ E) $\mathrm{E}=\frac{1}{2} \mathrm{mv}^{2}$ At the highest point vertical component of velocity is zero. Hence, $\mathrm{u}_{\mathrm{y}}=0$ and $\mathrm{u}_{\mathrm{x}}=\mathrm{v} \cos \theta$ Then, kinetic energy at highest point $\left(\mathrm{E}_{\mathrm{H}}\right)$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{~m}(\mathrm{v} \cos \theta)^{2} \quad \text { [From equation (i)] }$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{~m} \times\left(\mathrm{v} \cos 30^{\circ}\right)^{2}$ $\mathrm{E}_{\mathrm{H}} =\frac{1}{2} \mathrm{mv}^{2} \times\left(\frac{\sqrt{3}}{2}\right)^{2}$ $\mathrm{E}_{\mathrm{H}} =\frac{3 \mathrm{E}}{4}$