148917
A man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by $1 \mathrm{~ms}^{-1}$ and then has the same kinetic energy as the boy. The initial speed of the boy is
1 $\sqrt{2}+1 \mathrm{~ms}^{-1}$
2 $2(\sqrt{2}+1) \mathrm{ms}^{-1}$
3 $\sqrt{2} \mathrm{~ms}^{-1}$
4 $2 \mathrm{~ms}^{-1}$
Explanation:
B Given, Man boy Velocity $\mathrm{v}_{\mathrm{m}} \quad \mathrm{v}_{\mathrm{b}}$ Mass $m \quad m / 2$ K.E. of $\operatorname{man}=\frac{1}{2}[\mathrm{~K} . \mathrm{E}]_{\text {boy }} \quad$ (given) $\frac{1}{2} \mathrm{mv}_{\mathrm{m}}^{2} =\frac{1}{2}\left[\frac{1}{2} \times \frac{\mathrm{m}}{2} \times \mathrm{v}_{\mathrm{b}}^{2}\right]$ $\mathrm{v}_{\mathrm{m}}^{2} =\frac{\mathrm{v}_{\mathrm{b}}^{2}}{4}$ $\mathrm{v}_{\mathrm{m}} =\frac{\mathrm{v}_{\mathrm{b}}}{2}$ According to condition of question, we get the following equation- $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{m}}+1\right)^{2}=\frac{1}{2} \times \frac{\mathrm{m}}{2} \times \mathrm{v}_{\mathrm{b}}^{2}$ $\left(\mathrm{v}_{\mathrm{m}}+1\right)^{2}=\frac{\mathrm{v}_{\mathrm{b}}^{2}}{2}$ $\mathrm{v}_{\mathrm{m}}+1=\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}$ Putting the value of $\mathrm{v}_{\mathrm{m}}$ from equation (i), we get- $\frac{\mathrm{v}_{\mathrm{b}}}{2}+1=\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}$ $\frac{\mathrm{v}_{\mathrm{b}}}{2}-\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}=-1$ $-\left(\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}-\frac{\mathrm{v}_{\mathrm{b}}}{2}\right)=-1$ $\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}-\frac{\mathrm{v}_{\mathrm{b}}}{2}=1$ $\frac{\sqrt{2} \mathrm{v}_{\mathrm{b}}-\mathrm{v}_{\mathrm{b}}}{2}=1$ $\mathrm{v}_{\mathrm{b}}(\sqrt{2}-1)=2$ $\mathrm{v}_{\mathrm{b}}=\frac{2}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$ $\mathrm{v}_{\mathrm{b}}=\frac{2(\sqrt{2}+1)}{(\sqrt{2})^{2}-(1)^{2}}$ $\mathrm{v}_{\mathrm{b}}=2(\sqrt{2}+1) \mathrm{ms}^{-1}$
AP EAMCET (22.04.2019) Shift-I
Work, Energy and Power
148918
A girl in a swing is $2.5 \mathrm{~m}$ above ground at the maximum height and at $1.5 \mathrm{~m}$ above the ground at the lowest point. Her maximum velocity in the swing is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
1 $5 \sqrt{2} \mathrm{~ms}^{-1}$
2 $2 \sqrt{5} \mathrm{~ms}^{-1}$
3 $2 \sqrt{3} \mathrm{~ms}^{-1}$
4 $3 \sqrt{2} \mathrm{~ms}^{-1}$
5 $4 \sqrt{2} \mathrm{~ms}^{-1}$
Explanation:
B Given, maximum height $\left(\mathrm{h}_{1}\right)=2.5 \mathrm{~m}$, lowest height $\left(\mathrm{h}_{2}\right)=1.5 \mathrm{~m}$ As we know that, the maximum height, velocity (v) is zero. Then, total energy at maximum height- $\mathrm{E}_{1}=\mathrm{mgh}_{1}$ And total energy at lowest height- $\mathrm{E}_{2}=\mathrm{mgh}_{2}+\frac{1}{2} \mathrm{mv}_{\max }^{2}$ According to law of conservation of mechanical energy $\mathrm{E}_{1}=\mathrm{E}_{2}$ $\mathrm{mgh}_{1}=\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}+\mathrm{mgh}_{2}$ $\mathrm{mgh}_{1}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\text {max }}^{2}+2 \mathrm{gh}_{2}\right)$ $2 \mathrm{gh}_{1}=\mathrm{v}_{\text {max }}^{2}+2 \mathrm{gh}_{2}$ $\mathrm{v}_{\text {max }}^{2}=2 \mathrm{~g}\left(\mathrm{~h}_{1}-\mathrm{h}_{2}\right)$ $\mathrm{v}_{\max }=\sqrt{2 \times 10(2.5-1.5)}$ $\mathrm{v}_{\max }=2 \sqrt{5} \mathrm{~m} / \mathrm{s}$
Kerala CEE 2012
Work, Energy and Power
148919
The kinetic energy of a body becomes four times its initial value. The new momentum will be ...............
148917
A man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by $1 \mathrm{~ms}^{-1}$ and then has the same kinetic energy as the boy. The initial speed of the boy is
1 $\sqrt{2}+1 \mathrm{~ms}^{-1}$
2 $2(\sqrt{2}+1) \mathrm{ms}^{-1}$
3 $\sqrt{2} \mathrm{~ms}^{-1}$
4 $2 \mathrm{~ms}^{-1}$
Explanation:
B Given, Man boy Velocity $\mathrm{v}_{\mathrm{m}} \quad \mathrm{v}_{\mathrm{b}}$ Mass $m \quad m / 2$ K.E. of $\operatorname{man}=\frac{1}{2}[\mathrm{~K} . \mathrm{E}]_{\text {boy }} \quad$ (given) $\frac{1}{2} \mathrm{mv}_{\mathrm{m}}^{2} =\frac{1}{2}\left[\frac{1}{2} \times \frac{\mathrm{m}}{2} \times \mathrm{v}_{\mathrm{b}}^{2}\right]$ $\mathrm{v}_{\mathrm{m}}^{2} =\frac{\mathrm{v}_{\mathrm{b}}^{2}}{4}$ $\mathrm{v}_{\mathrm{m}} =\frac{\mathrm{v}_{\mathrm{b}}}{2}$ According to condition of question, we get the following equation- $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{m}}+1\right)^{2}=\frac{1}{2} \times \frac{\mathrm{m}}{2} \times \mathrm{v}_{\mathrm{b}}^{2}$ $\left(\mathrm{v}_{\mathrm{m}}+1\right)^{2}=\frac{\mathrm{v}_{\mathrm{b}}^{2}}{2}$ $\mathrm{v}_{\mathrm{m}}+1=\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}$ Putting the value of $\mathrm{v}_{\mathrm{m}}$ from equation (i), we get- $\frac{\mathrm{v}_{\mathrm{b}}}{2}+1=\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}$ $\frac{\mathrm{v}_{\mathrm{b}}}{2}-\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}=-1$ $-\left(\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}-\frac{\mathrm{v}_{\mathrm{b}}}{2}\right)=-1$ $\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}-\frac{\mathrm{v}_{\mathrm{b}}}{2}=1$ $\frac{\sqrt{2} \mathrm{v}_{\mathrm{b}}-\mathrm{v}_{\mathrm{b}}}{2}=1$ $\mathrm{v}_{\mathrm{b}}(\sqrt{2}-1)=2$ $\mathrm{v}_{\mathrm{b}}=\frac{2}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$ $\mathrm{v}_{\mathrm{b}}=\frac{2(\sqrt{2}+1)}{(\sqrt{2})^{2}-(1)^{2}}$ $\mathrm{v}_{\mathrm{b}}=2(\sqrt{2}+1) \mathrm{ms}^{-1}$
AP EAMCET (22.04.2019) Shift-I
Work, Energy and Power
148918
A girl in a swing is $2.5 \mathrm{~m}$ above ground at the maximum height and at $1.5 \mathrm{~m}$ above the ground at the lowest point. Her maximum velocity in the swing is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
1 $5 \sqrt{2} \mathrm{~ms}^{-1}$
2 $2 \sqrt{5} \mathrm{~ms}^{-1}$
3 $2 \sqrt{3} \mathrm{~ms}^{-1}$
4 $3 \sqrt{2} \mathrm{~ms}^{-1}$
5 $4 \sqrt{2} \mathrm{~ms}^{-1}$
Explanation:
B Given, maximum height $\left(\mathrm{h}_{1}\right)=2.5 \mathrm{~m}$, lowest height $\left(\mathrm{h}_{2}\right)=1.5 \mathrm{~m}$ As we know that, the maximum height, velocity (v) is zero. Then, total energy at maximum height- $\mathrm{E}_{1}=\mathrm{mgh}_{1}$ And total energy at lowest height- $\mathrm{E}_{2}=\mathrm{mgh}_{2}+\frac{1}{2} \mathrm{mv}_{\max }^{2}$ According to law of conservation of mechanical energy $\mathrm{E}_{1}=\mathrm{E}_{2}$ $\mathrm{mgh}_{1}=\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}+\mathrm{mgh}_{2}$ $\mathrm{mgh}_{1}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\text {max }}^{2}+2 \mathrm{gh}_{2}\right)$ $2 \mathrm{gh}_{1}=\mathrm{v}_{\text {max }}^{2}+2 \mathrm{gh}_{2}$ $\mathrm{v}_{\text {max }}^{2}=2 \mathrm{~g}\left(\mathrm{~h}_{1}-\mathrm{h}_{2}\right)$ $\mathrm{v}_{\max }=\sqrt{2 \times 10(2.5-1.5)}$ $\mathrm{v}_{\max }=2 \sqrt{5} \mathrm{~m} / \mathrm{s}$
Kerala CEE 2012
Work, Energy and Power
148919
The kinetic energy of a body becomes four times its initial value. The new momentum will be ...............
148917
A man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by $1 \mathrm{~ms}^{-1}$ and then has the same kinetic energy as the boy. The initial speed of the boy is
1 $\sqrt{2}+1 \mathrm{~ms}^{-1}$
2 $2(\sqrt{2}+1) \mathrm{ms}^{-1}$
3 $\sqrt{2} \mathrm{~ms}^{-1}$
4 $2 \mathrm{~ms}^{-1}$
Explanation:
B Given, Man boy Velocity $\mathrm{v}_{\mathrm{m}} \quad \mathrm{v}_{\mathrm{b}}$ Mass $m \quad m / 2$ K.E. of $\operatorname{man}=\frac{1}{2}[\mathrm{~K} . \mathrm{E}]_{\text {boy }} \quad$ (given) $\frac{1}{2} \mathrm{mv}_{\mathrm{m}}^{2} =\frac{1}{2}\left[\frac{1}{2} \times \frac{\mathrm{m}}{2} \times \mathrm{v}_{\mathrm{b}}^{2}\right]$ $\mathrm{v}_{\mathrm{m}}^{2} =\frac{\mathrm{v}_{\mathrm{b}}^{2}}{4}$ $\mathrm{v}_{\mathrm{m}} =\frac{\mathrm{v}_{\mathrm{b}}}{2}$ According to condition of question, we get the following equation- $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{m}}+1\right)^{2}=\frac{1}{2} \times \frac{\mathrm{m}}{2} \times \mathrm{v}_{\mathrm{b}}^{2}$ $\left(\mathrm{v}_{\mathrm{m}}+1\right)^{2}=\frac{\mathrm{v}_{\mathrm{b}}^{2}}{2}$ $\mathrm{v}_{\mathrm{m}}+1=\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}$ Putting the value of $\mathrm{v}_{\mathrm{m}}$ from equation (i), we get- $\frac{\mathrm{v}_{\mathrm{b}}}{2}+1=\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}$ $\frac{\mathrm{v}_{\mathrm{b}}}{2}-\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}=-1$ $-\left(\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}-\frac{\mathrm{v}_{\mathrm{b}}}{2}\right)=-1$ $\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}-\frac{\mathrm{v}_{\mathrm{b}}}{2}=1$ $\frac{\sqrt{2} \mathrm{v}_{\mathrm{b}}-\mathrm{v}_{\mathrm{b}}}{2}=1$ $\mathrm{v}_{\mathrm{b}}(\sqrt{2}-1)=2$ $\mathrm{v}_{\mathrm{b}}=\frac{2}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$ $\mathrm{v}_{\mathrm{b}}=\frac{2(\sqrt{2}+1)}{(\sqrt{2})^{2}-(1)^{2}}$ $\mathrm{v}_{\mathrm{b}}=2(\sqrt{2}+1) \mathrm{ms}^{-1}$
AP EAMCET (22.04.2019) Shift-I
Work, Energy and Power
148918
A girl in a swing is $2.5 \mathrm{~m}$ above ground at the maximum height and at $1.5 \mathrm{~m}$ above the ground at the lowest point. Her maximum velocity in the swing is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
1 $5 \sqrt{2} \mathrm{~ms}^{-1}$
2 $2 \sqrt{5} \mathrm{~ms}^{-1}$
3 $2 \sqrt{3} \mathrm{~ms}^{-1}$
4 $3 \sqrt{2} \mathrm{~ms}^{-1}$
5 $4 \sqrt{2} \mathrm{~ms}^{-1}$
Explanation:
B Given, maximum height $\left(\mathrm{h}_{1}\right)=2.5 \mathrm{~m}$, lowest height $\left(\mathrm{h}_{2}\right)=1.5 \mathrm{~m}$ As we know that, the maximum height, velocity (v) is zero. Then, total energy at maximum height- $\mathrm{E}_{1}=\mathrm{mgh}_{1}$ And total energy at lowest height- $\mathrm{E}_{2}=\mathrm{mgh}_{2}+\frac{1}{2} \mathrm{mv}_{\max }^{2}$ According to law of conservation of mechanical energy $\mathrm{E}_{1}=\mathrm{E}_{2}$ $\mathrm{mgh}_{1}=\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}+\mathrm{mgh}_{2}$ $\mathrm{mgh}_{1}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\text {max }}^{2}+2 \mathrm{gh}_{2}\right)$ $2 \mathrm{gh}_{1}=\mathrm{v}_{\text {max }}^{2}+2 \mathrm{gh}_{2}$ $\mathrm{v}_{\text {max }}^{2}=2 \mathrm{~g}\left(\mathrm{~h}_{1}-\mathrm{h}_{2}\right)$ $\mathrm{v}_{\max }=\sqrt{2 \times 10(2.5-1.5)}$ $\mathrm{v}_{\max }=2 \sqrt{5} \mathrm{~m} / \mathrm{s}$
Kerala CEE 2012
Work, Energy and Power
148919
The kinetic energy of a body becomes four times its initial value. The new momentum will be ...............
148917
A man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by $1 \mathrm{~ms}^{-1}$ and then has the same kinetic energy as the boy. The initial speed of the boy is
1 $\sqrt{2}+1 \mathrm{~ms}^{-1}$
2 $2(\sqrt{2}+1) \mathrm{ms}^{-1}$
3 $\sqrt{2} \mathrm{~ms}^{-1}$
4 $2 \mathrm{~ms}^{-1}$
Explanation:
B Given, Man boy Velocity $\mathrm{v}_{\mathrm{m}} \quad \mathrm{v}_{\mathrm{b}}$ Mass $m \quad m / 2$ K.E. of $\operatorname{man}=\frac{1}{2}[\mathrm{~K} . \mathrm{E}]_{\text {boy }} \quad$ (given) $\frac{1}{2} \mathrm{mv}_{\mathrm{m}}^{2} =\frac{1}{2}\left[\frac{1}{2} \times \frac{\mathrm{m}}{2} \times \mathrm{v}_{\mathrm{b}}^{2}\right]$ $\mathrm{v}_{\mathrm{m}}^{2} =\frac{\mathrm{v}_{\mathrm{b}}^{2}}{4}$ $\mathrm{v}_{\mathrm{m}} =\frac{\mathrm{v}_{\mathrm{b}}}{2}$ According to condition of question, we get the following equation- $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\mathrm{m}}+1\right)^{2}=\frac{1}{2} \times \frac{\mathrm{m}}{2} \times \mathrm{v}_{\mathrm{b}}^{2}$ $\left(\mathrm{v}_{\mathrm{m}}+1\right)^{2}=\frac{\mathrm{v}_{\mathrm{b}}^{2}}{2}$ $\mathrm{v}_{\mathrm{m}}+1=\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}$ Putting the value of $\mathrm{v}_{\mathrm{m}}$ from equation (i), we get- $\frac{\mathrm{v}_{\mathrm{b}}}{2}+1=\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}$ $\frac{\mathrm{v}_{\mathrm{b}}}{2}-\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}=-1$ $-\left(\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}-\frac{\mathrm{v}_{\mathrm{b}}}{2}\right)=-1$ $\frac{\mathrm{v}_{\mathrm{b}}}{\sqrt{2}}-\frac{\mathrm{v}_{\mathrm{b}}}{2}=1$ $\frac{\sqrt{2} \mathrm{v}_{\mathrm{b}}-\mathrm{v}_{\mathrm{b}}}{2}=1$ $\mathrm{v}_{\mathrm{b}}(\sqrt{2}-1)=2$ $\mathrm{v}_{\mathrm{b}}=\frac{2}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$ $\mathrm{v}_{\mathrm{b}}=\frac{2(\sqrt{2}+1)}{(\sqrt{2})^{2}-(1)^{2}}$ $\mathrm{v}_{\mathrm{b}}=2(\sqrt{2}+1) \mathrm{ms}^{-1}$
AP EAMCET (22.04.2019) Shift-I
Work, Energy and Power
148918
A girl in a swing is $2.5 \mathrm{~m}$ above ground at the maximum height and at $1.5 \mathrm{~m}$ above the ground at the lowest point. Her maximum velocity in the swing is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
1 $5 \sqrt{2} \mathrm{~ms}^{-1}$
2 $2 \sqrt{5} \mathrm{~ms}^{-1}$
3 $2 \sqrt{3} \mathrm{~ms}^{-1}$
4 $3 \sqrt{2} \mathrm{~ms}^{-1}$
5 $4 \sqrt{2} \mathrm{~ms}^{-1}$
Explanation:
B Given, maximum height $\left(\mathrm{h}_{1}\right)=2.5 \mathrm{~m}$, lowest height $\left(\mathrm{h}_{2}\right)=1.5 \mathrm{~m}$ As we know that, the maximum height, velocity (v) is zero. Then, total energy at maximum height- $\mathrm{E}_{1}=\mathrm{mgh}_{1}$ And total energy at lowest height- $\mathrm{E}_{2}=\mathrm{mgh}_{2}+\frac{1}{2} \mathrm{mv}_{\max }^{2}$ According to law of conservation of mechanical energy $\mathrm{E}_{1}=\mathrm{E}_{2}$ $\mathrm{mgh}_{1}=\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}+\mathrm{mgh}_{2}$ $\mathrm{mgh}_{1}=\frac{1}{2} \mathrm{~m}\left(\mathrm{v}_{\text {max }}^{2}+2 \mathrm{gh}_{2}\right)$ $2 \mathrm{gh}_{1}=\mathrm{v}_{\text {max }}^{2}+2 \mathrm{gh}_{2}$ $\mathrm{v}_{\text {max }}^{2}=2 \mathrm{~g}\left(\mathrm{~h}_{1}-\mathrm{h}_{2}\right)$ $\mathrm{v}_{\max }=\sqrt{2 \times 10(2.5-1.5)}$ $\mathrm{v}_{\max }=2 \sqrt{5} \mathrm{~m} / \mathrm{s}$
Kerala CEE 2012
Work, Energy and Power
148919
The kinetic energy of a body becomes four times its initial value. The new momentum will be ...............
