148966
Kinetic energy of a body increases times when its momentum is increased ' $n$ ' times:
1 $\mathrm{n}$
2 $2 \mathrm{n}$
3 $\sqrt{\mathrm{n}}$
4 $n^{2}$
Explanation:
D We know, Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{KE} =\frac{1}{2} \times \mathrm{mv}^{2} \times \frac{\mathrm{m}}{\mathrm{m}}$ $= \frac{1}{2} \times(\mathrm{mv})^{2} \times \frac{1}{\mathrm{~m}}$ $=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{KE} \propto \mathrm{p}^{2}$ So, if the momentum increased by $\mathrm{n}$ time [i.e. $\mathrm{p} \rightarrow \mathrm{np}$ ] $\mathrm{KE} \propto(n p)^{2}$ $\mathrm{KE} \propto n^{2} p^{2}$ Thus, if the momentum of a body is increased by $n$ times, then the kinetic energy also increases by $n^{2}$ times.
AP EAMCET-23.09.2020
Work, Energy and Power
148967
A uniform sphere of mass $m=2.5 \mathrm{~kg}$ and radius $R=10 \mathrm{~cm}$ rotates with an angular velocity $\omega=200 \mathrm{rad} / \mathrm{s}$ about its diameter. Find its kinetic energy.
1 $150 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $300 \mathrm{~J}$
4 $350 \mathrm{~J}$
Explanation:
B Given, Mass of sphere $(\mathrm{m})=2.5 \mathrm{~kg}$ Radius $(\mathrm{R})=10 \mathrm{~cm}=0.1 \mathrm{~m}$ Angular velocity $(\omega)=200 \mathrm{rad} / \mathrm{s}$. For sphere, $\mathrm{I}=\frac{2}{5} m R^{2}$ Rotational kinetic energy of Sphere (K.E.) $=\frac{1}{2} \mathrm{I} \omega^{2}$ Putting the value of I from equation (i), we get- $\text { K.E. }=\frac{1}{2} \times \frac{2}{5} \mathrm{mR}^{2} \omega^{2}$ $=\frac{1}{5} \times \mathrm{mR}^{2} \cdot \omega^{2}$ $=\frac{1}{5} \times 2.5 \times 0.1 \times 0.1 \times 200 \times 200$ $=\frac{25 \times 40}{5}$ Kinetic energy (K.E.) $=200 \mathrm{~J}$
TS EAMCET 29.09.2020
Work, Energy and Power
148968
A block of mass ' $m$ ' moving on a frictionless horizontal surface collides with a spring of spring constant ' $K$ ' and compresses it through a distance ' $x$ '. The maximum momentum of the block after collision is
1 $\sqrt{\mathrm{mK}} \mathrm{x}$
2 $\mathrm{mx}^{2} / \mathrm{K}$
3 Zero
4 $\mathrm{Kx}^{2} / 2 \mathrm{~m}$
Explanation:
A Momentum would be maximum when kinetic energy would be maximum. Total elastic potential energy of spring converted to kinetic energy. According to conservation of energy. $\mathrm{PE}_{\text {spring }}=\mathrm{KE}_{\text {mass }}$ $\frac{1}{2} \mathrm{Kx}^{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{mKx}^{2}=\mathrm{p}^{2}$ $\mathrm{p}=\sqrt{\mathrm{mKx}}$ $\mathrm{p}=\sqrt{\mathrm{mK}} \cdot \mathrm{x}$
MHT-CET 2020
Work, Energy and Power
148969
The kinetic energy of a light body and a heavy body is same. Which one of the following statements is CORRECT?
1 Body having high velocity has greater momentum
2 The heavy body has greater momentum
3 Both bodies have same momentum
4 The light body has greater momentum
Explanation:
B We know that, Kinetic energy (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ Or $\quad$ K.E. $=\frac{p^{2}}{2 m}$ $\mathrm{p}^{2}=2 \mathrm{~m} \text { (K.E.) }$ $\mathrm{p}=\sqrt{2 \mathrm{~m} \text { (K.E. })}$ $\because$ Momentum depends on mass $\mathrm{p} \propto \sqrt{\mathrm{m}}$ Kinetic energy for both bodies will be same but heavy body will have more momentum.
148966
Kinetic energy of a body increases times when its momentum is increased ' $n$ ' times:
1 $\mathrm{n}$
2 $2 \mathrm{n}$
3 $\sqrt{\mathrm{n}}$
4 $n^{2}$
Explanation:
D We know, Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{KE} =\frac{1}{2} \times \mathrm{mv}^{2} \times \frac{\mathrm{m}}{\mathrm{m}}$ $= \frac{1}{2} \times(\mathrm{mv})^{2} \times \frac{1}{\mathrm{~m}}$ $=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{KE} \propto \mathrm{p}^{2}$ So, if the momentum increased by $\mathrm{n}$ time [i.e. $\mathrm{p} \rightarrow \mathrm{np}$ ] $\mathrm{KE} \propto(n p)^{2}$ $\mathrm{KE} \propto n^{2} p^{2}$ Thus, if the momentum of a body is increased by $n$ times, then the kinetic energy also increases by $n^{2}$ times.
AP EAMCET-23.09.2020
Work, Energy and Power
148967
A uniform sphere of mass $m=2.5 \mathrm{~kg}$ and radius $R=10 \mathrm{~cm}$ rotates with an angular velocity $\omega=200 \mathrm{rad} / \mathrm{s}$ about its diameter. Find its kinetic energy.
1 $150 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $300 \mathrm{~J}$
4 $350 \mathrm{~J}$
Explanation:
B Given, Mass of sphere $(\mathrm{m})=2.5 \mathrm{~kg}$ Radius $(\mathrm{R})=10 \mathrm{~cm}=0.1 \mathrm{~m}$ Angular velocity $(\omega)=200 \mathrm{rad} / \mathrm{s}$. For sphere, $\mathrm{I}=\frac{2}{5} m R^{2}$ Rotational kinetic energy of Sphere (K.E.) $=\frac{1}{2} \mathrm{I} \omega^{2}$ Putting the value of I from equation (i), we get- $\text { K.E. }=\frac{1}{2} \times \frac{2}{5} \mathrm{mR}^{2} \omega^{2}$ $=\frac{1}{5} \times \mathrm{mR}^{2} \cdot \omega^{2}$ $=\frac{1}{5} \times 2.5 \times 0.1 \times 0.1 \times 200 \times 200$ $=\frac{25 \times 40}{5}$ Kinetic energy (K.E.) $=200 \mathrm{~J}$
TS EAMCET 29.09.2020
Work, Energy and Power
148968
A block of mass ' $m$ ' moving on a frictionless horizontal surface collides with a spring of spring constant ' $K$ ' and compresses it through a distance ' $x$ '. The maximum momentum of the block after collision is
1 $\sqrt{\mathrm{mK}} \mathrm{x}$
2 $\mathrm{mx}^{2} / \mathrm{K}$
3 Zero
4 $\mathrm{Kx}^{2} / 2 \mathrm{~m}$
Explanation:
A Momentum would be maximum when kinetic energy would be maximum. Total elastic potential energy of spring converted to kinetic energy. According to conservation of energy. $\mathrm{PE}_{\text {spring }}=\mathrm{KE}_{\text {mass }}$ $\frac{1}{2} \mathrm{Kx}^{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{mKx}^{2}=\mathrm{p}^{2}$ $\mathrm{p}=\sqrt{\mathrm{mKx}}$ $\mathrm{p}=\sqrt{\mathrm{mK}} \cdot \mathrm{x}$
MHT-CET 2020
Work, Energy and Power
148969
The kinetic energy of a light body and a heavy body is same. Which one of the following statements is CORRECT?
1 Body having high velocity has greater momentum
2 The heavy body has greater momentum
3 Both bodies have same momentum
4 The light body has greater momentum
Explanation:
B We know that, Kinetic energy (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ Or $\quad$ K.E. $=\frac{p^{2}}{2 m}$ $\mathrm{p}^{2}=2 \mathrm{~m} \text { (K.E.) }$ $\mathrm{p}=\sqrt{2 \mathrm{~m} \text { (K.E. })}$ $\because$ Momentum depends on mass $\mathrm{p} \propto \sqrt{\mathrm{m}}$ Kinetic energy for both bodies will be same but heavy body will have more momentum.
148966
Kinetic energy of a body increases times when its momentum is increased ' $n$ ' times:
1 $\mathrm{n}$
2 $2 \mathrm{n}$
3 $\sqrt{\mathrm{n}}$
4 $n^{2}$
Explanation:
D We know, Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{KE} =\frac{1}{2} \times \mathrm{mv}^{2} \times \frac{\mathrm{m}}{\mathrm{m}}$ $= \frac{1}{2} \times(\mathrm{mv})^{2} \times \frac{1}{\mathrm{~m}}$ $=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{KE} \propto \mathrm{p}^{2}$ So, if the momentum increased by $\mathrm{n}$ time [i.e. $\mathrm{p} \rightarrow \mathrm{np}$ ] $\mathrm{KE} \propto(n p)^{2}$ $\mathrm{KE} \propto n^{2} p^{2}$ Thus, if the momentum of a body is increased by $n$ times, then the kinetic energy also increases by $n^{2}$ times.
AP EAMCET-23.09.2020
Work, Energy and Power
148967
A uniform sphere of mass $m=2.5 \mathrm{~kg}$ and radius $R=10 \mathrm{~cm}$ rotates with an angular velocity $\omega=200 \mathrm{rad} / \mathrm{s}$ about its diameter. Find its kinetic energy.
1 $150 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $300 \mathrm{~J}$
4 $350 \mathrm{~J}$
Explanation:
B Given, Mass of sphere $(\mathrm{m})=2.5 \mathrm{~kg}$ Radius $(\mathrm{R})=10 \mathrm{~cm}=0.1 \mathrm{~m}$ Angular velocity $(\omega)=200 \mathrm{rad} / \mathrm{s}$. For sphere, $\mathrm{I}=\frac{2}{5} m R^{2}$ Rotational kinetic energy of Sphere (K.E.) $=\frac{1}{2} \mathrm{I} \omega^{2}$ Putting the value of I from equation (i), we get- $\text { K.E. }=\frac{1}{2} \times \frac{2}{5} \mathrm{mR}^{2} \omega^{2}$ $=\frac{1}{5} \times \mathrm{mR}^{2} \cdot \omega^{2}$ $=\frac{1}{5} \times 2.5 \times 0.1 \times 0.1 \times 200 \times 200$ $=\frac{25 \times 40}{5}$ Kinetic energy (K.E.) $=200 \mathrm{~J}$
TS EAMCET 29.09.2020
Work, Energy and Power
148968
A block of mass ' $m$ ' moving on a frictionless horizontal surface collides with a spring of spring constant ' $K$ ' and compresses it through a distance ' $x$ '. The maximum momentum of the block after collision is
1 $\sqrt{\mathrm{mK}} \mathrm{x}$
2 $\mathrm{mx}^{2} / \mathrm{K}$
3 Zero
4 $\mathrm{Kx}^{2} / 2 \mathrm{~m}$
Explanation:
A Momentum would be maximum when kinetic energy would be maximum. Total elastic potential energy of spring converted to kinetic energy. According to conservation of energy. $\mathrm{PE}_{\text {spring }}=\mathrm{KE}_{\text {mass }}$ $\frac{1}{2} \mathrm{Kx}^{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{mKx}^{2}=\mathrm{p}^{2}$ $\mathrm{p}=\sqrt{\mathrm{mKx}}$ $\mathrm{p}=\sqrt{\mathrm{mK}} \cdot \mathrm{x}$
MHT-CET 2020
Work, Energy and Power
148969
The kinetic energy of a light body and a heavy body is same. Which one of the following statements is CORRECT?
1 Body having high velocity has greater momentum
2 The heavy body has greater momentum
3 Both bodies have same momentum
4 The light body has greater momentum
Explanation:
B We know that, Kinetic energy (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ Or $\quad$ K.E. $=\frac{p^{2}}{2 m}$ $\mathrm{p}^{2}=2 \mathrm{~m} \text { (K.E.) }$ $\mathrm{p}=\sqrt{2 \mathrm{~m} \text { (K.E. })}$ $\because$ Momentum depends on mass $\mathrm{p} \propto \sqrt{\mathrm{m}}$ Kinetic energy for both bodies will be same but heavy body will have more momentum.
148966
Kinetic energy of a body increases times when its momentum is increased ' $n$ ' times:
1 $\mathrm{n}$
2 $2 \mathrm{n}$
3 $\sqrt{\mathrm{n}}$
4 $n^{2}$
Explanation:
D We know, Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{KE} =\frac{1}{2} \times \mathrm{mv}^{2} \times \frac{\mathrm{m}}{\mathrm{m}}$ $= \frac{1}{2} \times(\mathrm{mv})^{2} \times \frac{1}{\mathrm{~m}}$ $=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{KE} \propto \mathrm{p}^{2}$ So, if the momentum increased by $\mathrm{n}$ time [i.e. $\mathrm{p} \rightarrow \mathrm{np}$ ] $\mathrm{KE} \propto(n p)^{2}$ $\mathrm{KE} \propto n^{2} p^{2}$ Thus, if the momentum of a body is increased by $n$ times, then the kinetic energy also increases by $n^{2}$ times.
AP EAMCET-23.09.2020
Work, Energy and Power
148967
A uniform sphere of mass $m=2.5 \mathrm{~kg}$ and radius $R=10 \mathrm{~cm}$ rotates with an angular velocity $\omega=200 \mathrm{rad} / \mathrm{s}$ about its diameter. Find its kinetic energy.
1 $150 \mathrm{~J}$
2 $200 \mathrm{~J}$
3 $300 \mathrm{~J}$
4 $350 \mathrm{~J}$
Explanation:
B Given, Mass of sphere $(\mathrm{m})=2.5 \mathrm{~kg}$ Radius $(\mathrm{R})=10 \mathrm{~cm}=0.1 \mathrm{~m}$ Angular velocity $(\omega)=200 \mathrm{rad} / \mathrm{s}$. For sphere, $\mathrm{I}=\frac{2}{5} m R^{2}$ Rotational kinetic energy of Sphere (K.E.) $=\frac{1}{2} \mathrm{I} \omega^{2}$ Putting the value of I from equation (i), we get- $\text { K.E. }=\frac{1}{2} \times \frac{2}{5} \mathrm{mR}^{2} \omega^{2}$ $=\frac{1}{5} \times \mathrm{mR}^{2} \cdot \omega^{2}$ $=\frac{1}{5} \times 2.5 \times 0.1 \times 0.1 \times 200 \times 200$ $=\frac{25 \times 40}{5}$ Kinetic energy (K.E.) $=200 \mathrm{~J}$
TS EAMCET 29.09.2020
Work, Energy and Power
148968
A block of mass ' $m$ ' moving on a frictionless horizontal surface collides with a spring of spring constant ' $K$ ' and compresses it through a distance ' $x$ '. The maximum momentum of the block after collision is
1 $\sqrt{\mathrm{mK}} \mathrm{x}$
2 $\mathrm{mx}^{2} / \mathrm{K}$
3 Zero
4 $\mathrm{Kx}^{2} / 2 \mathrm{~m}$
Explanation:
A Momentum would be maximum when kinetic energy would be maximum. Total elastic potential energy of spring converted to kinetic energy. According to conservation of energy. $\mathrm{PE}_{\text {spring }}=\mathrm{KE}_{\text {mass }}$ $\frac{1}{2} \mathrm{Kx}^{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{mKx}^{2}=\mathrm{p}^{2}$ $\mathrm{p}=\sqrt{\mathrm{mKx}}$ $\mathrm{p}=\sqrt{\mathrm{mK}} \cdot \mathrm{x}$
MHT-CET 2020
Work, Energy and Power
148969
The kinetic energy of a light body and a heavy body is same. Which one of the following statements is CORRECT?
1 Body having high velocity has greater momentum
2 The heavy body has greater momentum
3 Both bodies have same momentum
4 The light body has greater momentum
Explanation:
B We know that, Kinetic energy (K.E.) $=\frac{1}{2} \mathrm{mv}^{2}$ Or $\quad$ K.E. $=\frac{p^{2}}{2 m}$ $\mathrm{p}^{2}=2 \mathrm{~m} \text { (K.E.) }$ $\mathrm{p}=\sqrt{2 \mathrm{~m} \text { (K.E. })}$ $\because$ Momentum depends on mass $\mathrm{p} \propto \sqrt{\mathrm{m}}$ Kinetic energy for both bodies will be same but heavy body will have more momentum.
