149135
$4 \mathrm{~m}^{3}$ of water is to be pumped to a height of $20 \mathrm{~m}$ and forced into a reservoir at a pressure of $2 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$. The work done by the motor is (external pressure $=10^{5} \mathrm{~N} / \mathrm{m}^{2}$ )
1 $8 \times 10^{5} \mathrm{~J}$
2 $16 \times 10^{5} \mathrm{~J}$
3 $12 \times 10^{5} \mathrm{~J}$
4 $32 \times 10^{5} \mathrm{~J}$
Explanation:
C Given that, Volume of pumped water $(\mathrm{v})=4 \mathrm{~m}^{3}$ Density of water $(\rho)=1 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ Gravitational acceleration $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$, Height of pumped water $(\mathrm{h})=20 \mathrm{~m}$ Here, $\quad \Delta \mathrm{p}=\left(2 \times 10^{5}-1 \times 10^{5}\right) \mathrm{N} / \mathrm{m}^{2}=1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$ Work done $=$ Change in gravitational potential energy + Work done against external pressure $\mathrm{W}=\mathrm{mgh}+\Delta \mathrm{p} \mathrm{v}$ $\mathrm{W} =\mathrm{v} \rho g h+\Delta \mathrm{p} \mathrm{v}$ $\therefore \quad \mathrm{W}=4 \times 1 \times 10^{3} \times 10 \times 20+1 \times 10^{5} \times 4$ $\mathrm{~W}=8 \times 10^{5}+4 \times 10^{5}$ $\mathrm{~W}=12 \times 10^{5} \mathrm{~J}$
BCECE-2008
Work, Energy and Power
149136
A meter stick is held vertically with one end on the floor and is then allowed to fall. Assuming that the end on the floor the stick does not slip, the velocity of the other end when it hits the floor, will be-
1 $10.8 \mathrm{~m} / \mathrm{s}$
2 $5.4 \mathrm{~m} / \mathrm{s}$
3 $2.5 \mathrm{~m} / \mathrm{s}$
4 none of these
Explanation:
B Let, $\text { Height of the meter stick }=\mathrm{L}$ $\text { Mass }=\mathrm{M}$ Center of mass meter stick fall from height $=\frac{1}{2} \mathrm{~L}$ According to law of conservation of energy $\operatorname{Mg} \frac{1}{2} \mathrm{~L}=\frac{1}{2} \mathrm{I} \omega^{2}$ Where, $I=\frac{1}{3} \mathrm{ML}^{2}, \quad$ and $\quad \omega=$ Angular velocity From the equation (i), $M g \frac{L}{2}=\frac{1}{2} \times \frac{M L^{2}}{3} \omega^{2}$ $M g \frac{L}{2}=\frac{M L^{2}}{6} \omega^{2}$ $\omega=\sqrt{\frac{3 g}{L}}$ Speed of other end, $v=\omega \mathrm{L}$ $\mathrm{v} =\omega=\sqrt{3 \mathrm{~g} / \mathrm{L}} \quad \text { [Where } \mathrm{L}=1 \mathrm{~m}]$ $\mathrm{v} =\sqrt{3 \mathrm{~g}}$ $\mathrm{v} =\sqrt{3 \times 9.8}$ $\mathrm{v} =\sqrt{29.4}$ $=5.4 \mathrm{~m} / \mathrm{s}$
BCECE-2007
Work, Energy and Power
149137
A body of mass $M$ moves with velocity $v$ and collides elastically with another body of mass $m$ $(M>>m)$ at rest, then the velocity of body of mass $m$ is :
1 $\mathrm{v}$
2 $2 \mathrm{v}$
3 $v / 2$
4 zero
Explanation:
B Applying conservation of momentum- $\mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}}$ $\mathrm{Mv}+\mathrm{m} \times 0=\mathrm{Mv}_{1}+\mathrm{mv}_{2}$ $\mathrm{Mv}=\mathrm{Mv}_{1}+\mathrm{mv}_{2}$ As given the collision is elastic, $\mathrm{e}=1$ $\mathrm{e}=1=\left(\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{u}_{1}-\mathrm{u}_{2}}\right)$ $1=\left(\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{v}-0}\right)$ $\Rightarrow \mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{v}$ $\Rightarrow \mathrm{v}_{1}=\mathrm{v}_{2}-\mathrm{v}$ On putting the value of equation (ii) in equation (i), we get $\mathrm{Mv}=M\left(\mathrm{v}_{2}-\mathrm{v}\right)+\mathrm{mv}_{2}$ $\mathrm{Mv}=\mathrm{Mv}_{2}-\mathrm{Mv}+\mathrm{mv}_{2}$ $2 \mathrm{Mv}=(\mathrm{M}+\mathrm{m}) \mathrm{v}_{2}$ $\mathrm{v}_{2}=\frac{2 \mathrm{Mv}}{\mathrm{M}+\mathrm{m}}$ According to question $\mathrm{M} \gg \gg \mathrm{m}$ $\text { So, } \mathrm{M}+\mathrm{m} \simeq \mathrm{M}$ $\therefore \mathrm{v}_{2}=\frac{2 \mathrm{Mv}}{\mathrm{M}}=2 \mathrm{v}$
BCECE-2004
Work, Energy and Power
149138
A spring is held compressed so that its stored energy is $2.4 \mathrm{~J}$. Its ends are in contact with masses $1 \mathrm{~g}$ and $48 \mathrm{~g}$ placed on a frictionless table. When the spring is released, the heavier mass will acquire a speed of-
1 $\frac{2.4}{49} \mathrm{~ms}^{-1}$
2 $\frac{2.4 \times 48}{49} \mathrm{~ms}^{-1}$
3 $\frac{10^{3}}{7} \mathrm{cms}^{-1}$
4 $\frac{10^{6}}{7} \mathrm{cms}^{-1}$
Explanation:
C Given, $\mathrm{m}_{1}=1 \mathrm{~g}, \mathrm{~m}_{2}=48 \mathrm{~g}$ Conservation of momentum, Now, $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $1 \times \mathrm{v}_{1}=48 \times \mathrm{v}_{2}$ $\mathrm{v}_{1}=48 \mathrm{v}_{2}$ According to law of conservation of energy, $\mathrm{PE}=\mathrm{KE}_{1}+\mathrm{KE}_{2}$ $2.4=\frac{1}{2} \mathrm{~m}_{1} \mathrm{v}_{1}^{2}+\frac{1}{2} \mathrm{~m}_{2} \mathrm{v}_{2}^{2}$ $2.4=\frac{1}{2} \times 1 \times 10^{-3} \times \mathrm{v}_{1}^{2}+\frac{1}{2} \times 48 \times 10^{-3} \times \mathrm{v}_{2}^{2}$ $2.4=\frac{1}{2} \times 10^{-3}\left[48^{2} \mathrm{v}_{2}^{2}+48 \mathrm{v}_{2}^{2}\right]$ $4.8 \times 10^{+3}=48 \mathrm{v}_{2}^{2}[49]$ $\mathrm{v}_{2}^{2}=\frac{100}{49}$ $\mathrm{v}_{2}=\frac{10}{7} \mathrm{~m} / \mathrm{s}$ $\mathrm{v}_{2}=\frac{10^{3}}{7} \mathrm{cms}^{-1}$
149135
$4 \mathrm{~m}^{3}$ of water is to be pumped to a height of $20 \mathrm{~m}$ and forced into a reservoir at a pressure of $2 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$. The work done by the motor is (external pressure $=10^{5} \mathrm{~N} / \mathrm{m}^{2}$ )
1 $8 \times 10^{5} \mathrm{~J}$
2 $16 \times 10^{5} \mathrm{~J}$
3 $12 \times 10^{5} \mathrm{~J}$
4 $32 \times 10^{5} \mathrm{~J}$
Explanation:
C Given that, Volume of pumped water $(\mathrm{v})=4 \mathrm{~m}^{3}$ Density of water $(\rho)=1 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ Gravitational acceleration $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$, Height of pumped water $(\mathrm{h})=20 \mathrm{~m}$ Here, $\quad \Delta \mathrm{p}=\left(2 \times 10^{5}-1 \times 10^{5}\right) \mathrm{N} / \mathrm{m}^{2}=1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$ Work done $=$ Change in gravitational potential energy + Work done against external pressure $\mathrm{W}=\mathrm{mgh}+\Delta \mathrm{p} \mathrm{v}$ $\mathrm{W} =\mathrm{v} \rho g h+\Delta \mathrm{p} \mathrm{v}$ $\therefore \quad \mathrm{W}=4 \times 1 \times 10^{3} \times 10 \times 20+1 \times 10^{5} \times 4$ $\mathrm{~W}=8 \times 10^{5}+4 \times 10^{5}$ $\mathrm{~W}=12 \times 10^{5} \mathrm{~J}$
BCECE-2008
Work, Energy and Power
149136
A meter stick is held vertically with one end on the floor and is then allowed to fall. Assuming that the end on the floor the stick does not slip, the velocity of the other end when it hits the floor, will be-
1 $10.8 \mathrm{~m} / \mathrm{s}$
2 $5.4 \mathrm{~m} / \mathrm{s}$
3 $2.5 \mathrm{~m} / \mathrm{s}$
4 none of these
Explanation:
B Let, $\text { Height of the meter stick }=\mathrm{L}$ $\text { Mass }=\mathrm{M}$ Center of mass meter stick fall from height $=\frac{1}{2} \mathrm{~L}$ According to law of conservation of energy $\operatorname{Mg} \frac{1}{2} \mathrm{~L}=\frac{1}{2} \mathrm{I} \omega^{2}$ Where, $I=\frac{1}{3} \mathrm{ML}^{2}, \quad$ and $\quad \omega=$ Angular velocity From the equation (i), $M g \frac{L}{2}=\frac{1}{2} \times \frac{M L^{2}}{3} \omega^{2}$ $M g \frac{L}{2}=\frac{M L^{2}}{6} \omega^{2}$ $\omega=\sqrt{\frac{3 g}{L}}$ Speed of other end, $v=\omega \mathrm{L}$ $\mathrm{v} =\omega=\sqrt{3 \mathrm{~g} / \mathrm{L}} \quad \text { [Where } \mathrm{L}=1 \mathrm{~m}]$ $\mathrm{v} =\sqrt{3 \mathrm{~g}}$ $\mathrm{v} =\sqrt{3 \times 9.8}$ $\mathrm{v} =\sqrt{29.4}$ $=5.4 \mathrm{~m} / \mathrm{s}$
BCECE-2007
Work, Energy and Power
149137
A body of mass $M$ moves with velocity $v$ and collides elastically with another body of mass $m$ $(M>>m)$ at rest, then the velocity of body of mass $m$ is :
1 $\mathrm{v}$
2 $2 \mathrm{v}$
3 $v / 2$
4 zero
Explanation:
B Applying conservation of momentum- $\mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}}$ $\mathrm{Mv}+\mathrm{m} \times 0=\mathrm{Mv}_{1}+\mathrm{mv}_{2}$ $\mathrm{Mv}=\mathrm{Mv}_{1}+\mathrm{mv}_{2}$ As given the collision is elastic, $\mathrm{e}=1$ $\mathrm{e}=1=\left(\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{u}_{1}-\mathrm{u}_{2}}\right)$ $1=\left(\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{v}-0}\right)$ $\Rightarrow \mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{v}$ $\Rightarrow \mathrm{v}_{1}=\mathrm{v}_{2}-\mathrm{v}$ On putting the value of equation (ii) in equation (i), we get $\mathrm{Mv}=M\left(\mathrm{v}_{2}-\mathrm{v}\right)+\mathrm{mv}_{2}$ $\mathrm{Mv}=\mathrm{Mv}_{2}-\mathrm{Mv}+\mathrm{mv}_{2}$ $2 \mathrm{Mv}=(\mathrm{M}+\mathrm{m}) \mathrm{v}_{2}$ $\mathrm{v}_{2}=\frac{2 \mathrm{Mv}}{\mathrm{M}+\mathrm{m}}$ According to question $\mathrm{M} \gg \gg \mathrm{m}$ $\text { So, } \mathrm{M}+\mathrm{m} \simeq \mathrm{M}$ $\therefore \mathrm{v}_{2}=\frac{2 \mathrm{Mv}}{\mathrm{M}}=2 \mathrm{v}$
BCECE-2004
Work, Energy and Power
149138
A spring is held compressed so that its stored energy is $2.4 \mathrm{~J}$. Its ends are in contact with masses $1 \mathrm{~g}$ and $48 \mathrm{~g}$ placed on a frictionless table. When the spring is released, the heavier mass will acquire a speed of-
1 $\frac{2.4}{49} \mathrm{~ms}^{-1}$
2 $\frac{2.4 \times 48}{49} \mathrm{~ms}^{-1}$
3 $\frac{10^{3}}{7} \mathrm{cms}^{-1}$
4 $\frac{10^{6}}{7} \mathrm{cms}^{-1}$
Explanation:
C Given, $\mathrm{m}_{1}=1 \mathrm{~g}, \mathrm{~m}_{2}=48 \mathrm{~g}$ Conservation of momentum, Now, $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $1 \times \mathrm{v}_{1}=48 \times \mathrm{v}_{2}$ $\mathrm{v}_{1}=48 \mathrm{v}_{2}$ According to law of conservation of energy, $\mathrm{PE}=\mathrm{KE}_{1}+\mathrm{KE}_{2}$ $2.4=\frac{1}{2} \mathrm{~m}_{1} \mathrm{v}_{1}^{2}+\frac{1}{2} \mathrm{~m}_{2} \mathrm{v}_{2}^{2}$ $2.4=\frac{1}{2} \times 1 \times 10^{-3} \times \mathrm{v}_{1}^{2}+\frac{1}{2} \times 48 \times 10^{-3} \times \mathrm{v}_{2}^{2}$ $2.4=\frac{1}{2} \times 10^{-3}\left[48^{2} \mathrm{v}_{2}^{2}+48 \mathrm{v}_{2}^{2}\right]$ $4.8 \times 10^{+3}=48 \mathrm{v}_{2}^{2}[49]$ $\mathrm{v}_{2}^{2}=\frac{100}{49}$ $\mathrm{v}_{2}=\frac{10}{7} \mathrm{~m} / \mathrm{s}$ $\mathrm{v}_{2}=\frac{10^{3}}{7} \mathrm{cms}^{-1}$
149135
$4 \mathrm{~m}^{3}$ of water is to be pumped to a height of $20 \mathrm{~m}$ and forced into a reservoir at a pressure of $2 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$. The work done by the motor is (external pressure $=10^{5} \mathrm{~N} / \mathrm{m}^{2}$ )
1 $8 \times 10^{5} \mathrm{~J}$
2 $16 \times 10^{5} \mathrm{~J}$
3 $12 \times 10^{5} \mathrm{~J}$
4 $32 \times 10^{5} \mathrm{~J}$
Explanation:
C Given that, Volume of pumped water $(\mathrm{v})=4 \mathrm{~m}^{3}$ Density of water $(\rho)=1 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ Gravitational acceleration $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$, Height of pumped water $(\mathrm{h})=20 \mathrm{~m}$ Here, $\quad \Delta \mathrm{p}=\left(2 \times 10^{5}-1 \times 10^{5}\right) \mathrm{N} / \mathrm{m}^{2}=1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$ Work done $=$ Change in gravitational potential energy + Work done against external pressure $\mathrm{W}=\mathrm{mgh}+\Delta \mathrm{p} \mathrm{v}$ $\mathrm{W} =\mathrm{v} \rho g h+\Delta \mathrm{p} \mathrm{v}$ $\therefore \quad \mathrm{W}=4 \times 1 \times 10^{3} \times 10 \times 20+1 \times 10^{5} \times 4$ $\mathrm{~W}=8 \times 10^{5}+4 \times 10^{5}$ $\mathrm{~W}=12 \times 10^{5} \mathrm{~J}$
BCECE-2008
Work, Energy and Power
149136
A meter stick is held vertically with one end on the floor and is then allowed to fall. Assuming that the end on the floor the stick does not slip, the velocity of the other end when it hits the floor, will be-
1 $10.8 \mathrm{~m} / \mathrm{s}$
2 $5.4 \mathrm{~m} / \mathrm{s}$
3 $2.5 \mathrm{~m} / \mathrm{s}$
4 none of these
Explanation:
B Let, $\text { Height of the meter stick }=\mathrm{L}$ $\text { Mass }=\mathrm{M}$ Center of mass meter stick fall from height $=\frac{1}{2} \mathrm{~L}$ According to law of conservation of energy $\operatorname{Mg} \frac{1}{2} \mathrm{~L}=\frac{1}{2} \mathrm{I} \omega^{2}$ Where, $I=\frac{1}{3} \mathrm{ML}^{2}, \quad$ and $\quad \omega=$ Angular velocity From the equation (i), $M g \frac{L}{2}=\frac{1}{2} \times \frac{M L^{2}}{3} \omega^{2}$ $M g \frac{L}{2}=\frac{M L^{2}}{6} \omega^{2}$ $\omega=\sqrt{\frac{3 g}{L}}$ Speed of other end, $v=\omega \mathrm{L}$ $\mathrm{v} =\omega=\sqrt{3 \mathrm{~g} / \mathrm{L}} \quad \text { [Where } \mathrm{L}=1 \mathrm{~m}]$ $\mathrm{v} =\sqrt{3 \mathrm{~g}}$ $\mathrm{v} =\sqrt{3 \times 9.8}$ $\mathrm{v} =\sqrt{29.4}$ $=5.4 \mathrm{~m} / \mathrm{s}$
BCECE-2007
Work, Energy and Power
149137
A body of mass $M$ moves with velocity $v$ and collides elastically with another body of mass $m$ $(M>>m)$ at rest, then the velocity of body of mass $m$ is :
1 $\mathrm{v}$
2 $2 \mathrm{v}$
3 $v / 2$
4 zero
Explanation:
B Applying conservation of momentum- $\mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}}$ $\mathrm{Mv}+\mathrm{m} \times 0=\mathrm{Mv}_{1}+\mathrm{mv}_{2}$ $\mathrm{Mv}=\mathrm{Mv}_{1}+\mathrm{mv}_{2}$ As given the collision is elastic, $\mathrm{e}=1$ $\mathrm{e}=1=\left(\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{u}_{1}-\mathrm{u}_{2}}\right)$ $1=\left(\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{v}-0}\right)$ $\Rightarrow \mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{v}$ $\Rightarrow \mathrm{v}_{1}=\mathrm{v}_{2}-\mathrm{v}$ On putting the value of equation (ii) in equation (i), we get $\mathrm{Mv}=M\left(\mathrm{v}_{2}-\mathrm{v}\right)+\mathrm{mv}_{2}$ $\mathrm{Mv}=\mathrm{Mv}_{2}-\mathrm{Mv}+\mathrm{mv}_{2}$ $2 \mathrm{Mv}=(\mathrm{M}+\mathrm{m}) \mathrm{v}_{2}$ $\mathrm{v}_{2}=\frac{2 \mathrm{Mv}}{\mathrm{M}+\mathrm{m}}$ According to question $\mathrm{M} \gg \gg \mathrm{m}$ $\text { So, } \mathrm{M}+\mathrm{m} \simeq \mathrm{M}$ $\therefore \mathrm{v}_{2}=\frac{2 \mathrm{Mv}}{\mathrm{M}}=2 \mathrm{v}$
BCECE-2004
Work, Energy and Power
149138
A spring is held compressed so that its stored energy is $2.4 \mathrm{~J}$. Its ends are in contact with masses $1 \mathrm{~g}$ and $48 \mathrm{~g}$ placed on a frictionless table. When the spring is released, the heavier mass will acquire a speed of-
1 $\frac{2.4}{49} \mathrm{~ms}^{-1}$
2 $\frac{2.4 \times 48}{49} \mathrm{~ms}^{-1}$
3 $\frac{10^{3}}{7} \mathrm{cms}^{-1}$
4 $\frac{10^{6}}{7} \mathrm{cms}^{-1}$
Explanation:
C Given, $\mathrm{m}_{1}=1 \mathrm{~g}, \mathrm{~m}_{2}=48 \mathrm{~g}$ Conservation of momentum, Now, $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $1 \times \mathrm{v}_{1}=48 \times \mathrm{v}_{2}$ $\mathrm{v}_{1}=48 \mathrm{v}_{2}$ According to law of conservation of energy, $\mathrm{PE}=\mathrm{KE}_{1}+\mathrm{KE}_{2}$ $2.4=\frac{1}{2} \mathrm{~m}_{1} \mathrm{v}_{1}^{2}+\frac{1}{2} \mathrm{~m}_{2} \mathrm{v}_{2}^{2}$ $2.4=\frac{1}{2} \times 1 \times 10^{-3} \times \mathrm{v}_{1}^{2}+\frac{1}{2} \times 48 \times 10^{-3} \times \mathrm{v}_{2}^{2}$ $2.4=\frac{1}{2} \times 10^{-3}\left[48^{2} \mathrm{v}_{2}^{2}+48 \mathrm{v}_{2}^{2}\right]$ $4.8 \times 10^{+3}=48 \mathrm{v}_{2}^{2}[49]$ $\mathrm{v}_{2}^{2}=\frac{100}{49}$ $\mathrm{v}_{2}=\frac{10}{7} \mathrm{~m} / \mathrm{s}$ $\mathrm{v}_{2}=\frac{10^{3}}{7} \mathrm{cms}^{-1}$
149135
$4 \mathrm{~m}^{3}$ of water is to be pumped to a height of $20 \mathrm{~m}$ and forced into a reservoir at a pressure of $2 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$. The work done by the motor is (external pressure $=10^{5} \mathrm{~N} / \mathrm{m}^{2}$ )
1 $8 \times 10^{5} \mathrm{~J}$
2 $16 \times 10^{5} \mathrm{~J}$
3 $12 \times 10^{5} \mathrm{~J}$
4 $32 \times 10^{5} \mathrm{~J}$
Explanation:
C Given that, Volume of pumped water $(\mathrm{v})=4 \mathrm{~m}^{3}$ Density of water $(\rho)=1 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}$ Gravitational acceleration $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$, Height of pumped water $(\mathrm{h})=20 \mathrm{~m}$ Here, $\quad \Delta \mathrm{p}=\left(2 \times 10^{5}-1 \times 10^{5}\right) \mathrm{N} / \mathrm{m}^{2}=1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}$ Work done $=$ Change in gravitational potential energy + Work done against external pressure $\mathrm{W}=\mathrm{mgh}+\Delta \mathrm{p} \mathrm{v}$ $\mathrm{W} =\mathrm{v} \rho g h+\Delta \mathrm{p} \mathrm{v}$ $\therefore \quad \mathrm{W}=4 \times 1 \times 10^{3} \times 10 \times 20+1 \times 10^{5} \times 4$ $\mathrm{~W}=8 \times 10^{5}+4 \times 10^{5}$ $\mathrm{~W}=12 \times 10^{5} \mathrm{~J}$
BCECE-2008
Work, Energy and Power
149136
A meter stick is held vertically with one end on the floor and is then allowed to fall. Assuming that the end on the floor the stick does not slip, the velocity of the other end when it hits the floor, will be-
1 $10.8 \mathrm{~m} / \mathrm{s}$
2 $5.4 \mathrm{~m} / \mathrm{s}$
3 $2.5 \mathrm{~m} / \mathrm{s}$
4 none of these
Explanation:
B Let, $\text { Height of the meter stick }=\mathrm{L}$ $\text { Mass }=\mathrm{M}$ Center of mass meter stick fall from height $=\frac{1}{2} \mathrm{~L}$ According to law of conservation of energy $\operatorname{Mg} \frac{1}{2} \mathrm{~L}=\frac{1}{2} \mathrm{I} \omega^{2}$ Where, $I=\frac{1}{3} \mathrm{ML}^{2}, \quad$ and $\quad \omega=$ Angular velocity From the equation (i), $M g \frac{L}{2}=\frac{1}{2} \times \frac{M L^{2}}{3} \omega^{2}$ $M g \frac{L}{2}=\frac{M L^{2}}{6} \omega^{2}$ $\omega=\sqrt{\frac{3 g}{L}}$ Speed of other end, $v=\omega \mathrm{L}$ $\mathrm{v} =\omega=\sqrt{3 \mathrm{~g} / \mathrm{L}} \quad \text { [Where } \mathrm{L}=1 \mathrm{~m}]$ $\mathrm{v} =\sqrt{3 \mathrm{~g}}$ $\mathrm{v} =\sqrt{3 \times 9.8}$ $\mathrm{v} =\sqrt{29.4}$ $=5.4 \mathrm{~m} / \mathrm{s}$
BCECE-2007
Work, Energy and Power
149137
A body of mass $M$ moves with velocity $v$ and collides elastically with another body of mass $m$ $(M>>m)$ at rest, then the velocity of body of mass $m$ is :
1 $\mathrm{v}$
2 $2 \mathrm{v}$
3 $v / 2$
4 zero
Explanation:
B Applying conservation of momentum- $\mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}}$ $\mathrm{Mv}+\mathrm{m} \times 0=\mathrm{Mv}_{1}+\mathrm{mv}_{2}$ $\mathrm{Mv}=\mathrm{Mv}_{1}+\mathrm{mv}_{2}$ As given the collision is elastic, $\mathrm{e}=1$ $\mathrm{e}=1=\left(\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{u}_{1}-\mathrm{u}_{2}}\right)$ $1=\left(\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{v}-0}\right)$ $\Rightarrow \mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{v}$ $\Rightarrow \mathrm{v}_{1}=\mathrm{v}_{2}-\mathrm{v}$ On putting the value of equation (ii) in equation (i), we get $\mathrm{Mv}=M\left(\mathrm{v}_{2}-\mathrm{v}\right)+\mathrm{mv}_{2}$ $\mathrm{Mv}=\mathrm{Mv}_{2}-\mathrm{Mv}+\mathrm{mv}_{2}$ $2 \mathrm{Mv}=(\mathrm{M}+\mathrm{m}) \mathrm{v}_{2}$ $\mathrm{v}_{2}=\frac{2 \mathrm{Mv}}{\mathrm{M}+\mathrm{m}}$ According to question $\mathrm{M} \gg \gg \mathrm{m}$ $\text { So, } \mathrm{M}+\mathrm{m} \simeq \mathrm{M}$ $\therefore \mathrm{v}_{2}=\frac{2 \mathrm{Mv}}{\mathrm{M}}=2 \mathrm{v}$
BCECE-2004
Work, Energy and Power
149138
A spring is held compressed so that its stored energy is $2.4 \mathrm{~J}$. Its ends are in contact with masses $1 \mathrm{~g}$ and $48 \mathrm{~g}$ placed on a frictionless table. When the spring is released, the heavier mass will acquire a speed of-
1 $\frac{2.4}{49} \mathrm{~ms}^{-1}$
2 $\frac{2.4 \times 48}{49} \mathrm{~ms}^{-1}$
3 $\frac{10^{3}}{7} \mathrm{cms}^{-1}$
4 $\frac{10^{6}}{7} \mathrm{cms}^{-1}$
Explanation:
C Given, $\mathrm{m}_{1}=1 \mathrm{~g}, \mathrm{~m}_{2}=48 \mathrm{~g}$ Conservation of momentum, Now, $\mathrm{m}_{1} \mathrm{v}_{1}=\mathrm{m}_{2} \mathrm{v}_{2}$ $1 \times \mathrm{v}_{1}=48 \times \mathrm{v}_{2}$ $\mathrm{v}_{1}=48 \mathrm{v}_{2}$ According to law of conservation of energy, $\mathrm{PE}=\mathrm{KE}_{1}+\mathrm{KE}_{2}$ $2.4=\frac{1}{2} \mathrm{~m}_{1} \mathrm{v}_{1}^{2}+\frac{1}{2} \mathrm{~m}_{2} \mathrm{v}_{2}^{2}$ $2.4=\frac{1}{2} \times 1 \times 10^{-3} \times \mathrm{v}_{1}^{2}+\frac{1}{2} \times 48 \times 10^{-3} \times \mathrm{v}_{2}^{2}$ $2.4=\frac{1}{2} \times 10^{-3}\left[48^{2} \mathrm{v}_{2}^{2}+48 \mathrm{v}_{2}^{2}\right]$ $4.8 \times 10^{+3}=48 \mathrm{v}_{2}^{2}[49]$ $\mathrm{v}_{2}^{2}=\frac{100}{49}$ $\mathrm{v}_{2}=\frac{10}{7} \mathrm{~m} / \mathrm{s}$ $\mathrm{v}_{2}=\frac{10^{3}}{7} \mathrm{cms}^{-1}$
