146350
A rocket with an initial mass \(m_{0}\) is going up with a constant acceleration a by exhausting gases with a velocity \(v\) relative to the rocket motion, then the mass of the rocket at any instant of time is (assume that no other forces act on it)
1 \(m=m_{0} e^{-\frac{a t}{v}}\)
2 \(m=m_{0} e^{-\frac{2 a t}{v}}\)
3 \(m=m_{0} e^{-\frac{a t}{2 v}}\)
4 \(m=m_{0} e^{-\frac{a^{2} t^{2}}{v^{2}}}\)
Explanation:
A Given, External forces on rocket is zero \(\left(\mathrm{F}_{\text {ext }}=0\right)\). External force \(=\) Rate of change in momentum \(F_{\text {ext }}=\frac{d p}{d t}=\frac{d(m v)}{d t}\) \(F_{\text {ext }}=m \frac{d v}{d t}+v \frac{d m}{d t}\) \(m \frac{d v}{d t}+v \frac{d m}{d t}=0\) \(\left(\text { Where } \frac{d v}{d t}=a\right)\) \(\therefore \quad m a=-v \frac{d m}{d t}\) \(\frac{-a}{v} d t=\frac{d m}{m}\) Integrating both side \(\frac{-\mathrm{a}}{\mathrm{v}} \int_{0}^{\mathrm{t}} \mathrm{dt}=\int_{\mathrm{m}_{0}}^{\mathrm{m}} \frac{\mathrm{dm}}{\mathrm{m}}\) \(-\frac{\mathrm{a}}{\mathrm{v}}[\mathrm{t}]_{0}^{\mathrm{t}}=\left[\log _{\mathrm{e}} \mathrm{m}\right]_{\mathrm{m}_{0}}^{\mathrm{m}}\) \(\frac{-\mathrm{at}}{\mathrm{v}}=\log _{\mathrm{e}} \frac{\mathrm{m}}{\mathrm{m}_{0}}\) \(\frac{\mathrm{m}}{\mathrm{m}_{0}}=\mathrm{e}^{-\mathrm{at} / \mathrm{v}}\) \(\mathrm{m}=\mathrm{m}_{0} \mathrm{e}^{-\mathrm{at} / \mathrm{v}}\)
AP EAMCET (23.04.2018) Shift-2
Laws of Motion
146352
A \(600 \mathrm{~kg}\) rocket is set for a vertical firing. If the exhaust speed is \(1000 \mathrm{~ms}^{-1}\), the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is
1 \(117.6 \mathrm{~kg} \mathrm{~s}^{-1}\)
2 \(58.6 \mathrm{~kg} \mathrm{~s}^{-1}\)
3 \(6 \mathrm{~kg} \mathrm{~s}^{-1}\)
4 \(76.4 \mathrm{~kg} \mathrm{~s}^{-1}\)
Explanation:
C Given, \(\mathrm{m}=600 \mathrm{~kg}\) Force required to overcome the weight of rocket is, \(\mathrm{F}=\mathrm{mg}\) \(\frac{\mathrm{dp}}{\mathrm{dt}}=\mathrm{mg}\) \(\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}}=\mathrm{mg}\) \(\mathrm{v} \frac{\mathrm{dm}}{\mathrm{dt}}=\mathrm{mg}\) \(\frac{\mathrm{dm}}{\mathrm{dt}}=\frac{\mathrm{mg}}{\mathrm{v}}\) \(\frac{\mathrm{dm}}{\mathrm{dt}}=\frac{600 \times 10}{1000}\) \(\frac{\mathrm{dm}}{\mathrm{dt}}=6 \mathrm{kgs}^{-1}\) \(\left[\because \frac{\mathrm{dv}}{\mathrm{dt}}=0\right]\)
AIPMT- 1990
Laws of Motion
146353
If the force on a rocket moving with a velocity of \(300 \mathrm{~m} / \mathrm{s}\) is \(345 \mathrm{~N}\), then the rate of combustion of the fuel is
1 \(0.55 \mathrm{~kg} / \mathrm{s}\)
2 \(0.75 \mathrm{~kg} / \mathrm{s}\)
3 \(1.15 \mathrm{~kg} / \mathrm{s}\)
4 \(2.25 \mathrm{~kg} / \mathrm{s}\)
Explanation:
C Given that, \(\mathrm{v}=300 \mathrm{~m} / \mathrm{s}\) \(\mathrm{F}=345 \mathrm{~N}\). \(\frac{\mathrm{dm}}{\mathrm{dt}}=\) ? \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}}=\) Rate of change in momentum \(\mathrm{F} =\mathrm{v} \frac{\mathrm{dm}}{\mathrm{dt}}\) \(345 =300 \frac{\mathrm{dm}}{\mathrm{dt}}\) \(\frac{\mathrm{dm}}{\mathrm{dt}} =\frac{345}{300}\) \(\frac{\mathrm{dm}}{\mathrm{dt}} =1.15 \mathrm{~kg} / \mathrm{s}\)
AIPMT- 1995
Laws of Motion
146354
A satellite in a force free space sweeps stationary interplanetary dust at a rate. \(\left(\frac{\mathbf{d M}}{\mathbf{d t}}\right)=\alpha \mathrm{v}\). The acceleration of satellite is
1 \(-\frac{2 \alpha v^{2}}{M}\)
2 \(-\frac{\alpha v^{2}}{M}\)
3 \(-\frac{\alpha v^{2}}{2 M}\)
4 \(-\alpha v^{2}\)
Explanation:
B Given, \(\frac{\mathrm{dM}}{\mathrm{dt}}=\alpha \mathrm{v}\) From, Newton's second law of motion \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(F=\frac{d}{d t}(M v)\) \(F=M \frac{d v}{d t}+v \frac{d M}{d t}\) Given the space is force free then net force is zero \((F=\) \(0)\), \(0=\mathrm{M} \frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{v} \frac{\mathrm{dM}}{\mathrm{dt}}\) We know that, \(\quad a=\frac{d v}{d t}\) From equation (i), \(0=\mathrm{Ma}+\mathrm{v}(\alpha \mathrm{v})\) \(-\mathrm{Ma}=\alpha \mathrm{v}^{2}\) \(\mathrm{a}=-\frac{\alpha \cdot \mathrm{v}^{2}}{\mathrm{M}}\)
146350
A rocket with an initial mass \(m_{0}\) is going up with a constant acceleration a by exhausting gases with a velocity \(v\) relative to the rocket motion, then the mass of the rocket at any instant of time is (assume that no other forces act on it)
1 \(m=m_{0} e^{-\frac{a t}{v}}\)
2 \(m=m_{0} e^{-\frac{2 a t}{v}}\)
3 \(m=m_{0} e^{-\frac{a t}{2 v}}\)
4 \(m=m_{0} e^{-\frac{a^{2} t^{2}}{v^{2}}}\)
Explanation:
A Given, External forces on rocket is zero \(\left(\mathrm{F}_{\text {ext }}=0\right)\). External force \(=\) Rate of change in momentum \(F_{\text {ext }}=\frac{d p}{d t}=\frac{d(m v)}{d t}\) \(F_{\text {ext }}=m \frac{d v}{d t}+v \frac{d m}{d t}\) \(m \frac{d v}{d t}+v \frac{d m}{d t}=0\) \(\left(\text { Where } \frac{d v}{d t}=a\right)\) \(\therefore \quad m a=-v \frac{d m}{d t}\) \(\frac{-a}{v} d t=\frac{d m}{m}\) Integrating both side \(\frac{-\mathrm{a}}{\mathrm{v}} \int_{0}^{\mathrm{t}} \mathrm{dt}=\int_{\mathrm{m}_{0}}^{\mathrm{m}} \frac{\mathrm{dm}}{\mathrm{m}}\) \(-\frac{\mathrm{a}}{\mathrm{v}}[\mathrm{t}]_{0}^{\mathrm{t}}=\left[\log _{\mathrm{e}} \mathrm{m}\right]_{\mathrm{m}_{0}}^{\mathrm{m}}\) \(\frac{-\mathrm{at}}{\mathrm{v}}=\log _{\mathrm{e}} \frac{\mathrm{m}}{\mathrm{m}_{0}}\) \(\frac{\mathrm{m}}{\mathrm{m}_{0}}=\mathrm{e}^{-\mathrm{at} / \mathrm{v}}\) \(\mathrm{m}=\mathrm{m}_{0} \mathrm{e}^{-\mathrm{at} / \mathrm{v}}\)
AP EAMCET (23.04.2018) Shift-2
Laws of Motion
146352
A \(600 \mathrm{~kg}\) rocket is set for a vertical firing. If the exhaust speed is \(1000 \mathrm{~ms}^{-1}\), the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is
1 \(117.6 \mathrm{~kg} \mathrm{~s}^{-1}\)
2 \(58.6 \mathrm{~kg} \mathrm{~s}^{-1}\)
3 \(6 \mathrm{~kg} \mathrm{~s}^{-1}\)
4 \(76.4 \mathrm{~kg} \mathrm{~s}^{-1}\)
Explanation:
C Given, \(\mathrm{m}=600 \mathrm{~kg}\) Force required to overcome the weight of rocket is, \(\mathrm{F}=\mathrm{mg}\) \(\frac{\mathrm{dp}}{\mathrm{dt}}=\mathrm{mg}\) \(\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}}=\mathrm{mg}\) \(\mathrm{v} \frac{\mathrm{dm}}{\mathrm{dt}}=\mathrm{mg}\) \(\frac{\mathrm{dm}}{\mathrm{dt}}=\frac{\mathrm{mg}}{\mathrm{v}}\) \(\frac{\mathrm{dm}}{\mathrm{dt}}=\frac{600 \times 10}{1000}\) \(\frac{\mathrm{dm}}{\mathrm{dt}}=6 \mathrm{kgs}^{-1}\) \(\left[\because \frac{\mathrm{dv}}{\mathrm{dt}}=0\right]\)
AIPMT- 1990
Laws of Motion
146353
If the force on a rocket moving with a velocity of \(300 \mathrm{~m} / \mathrm{s}\) is \(345 \mathrm{~N}\), then the rate of combustion of the fuel is
1 \(0.55 \mathrm{~kg} / \mathrm{s}\)
2 \(0.75 \mathrm{~kg} / \mathrm{s}\)
3 \(1.15 \mathrm{~kg} / \mathrm{s}\)
4 \(2.25 \mathrm{~kg} / \mathrm{s}\)
Explanation:
C Given that, \(\mathrm{v}=300 \mathrm{~m} / \mathrm{s}\) \(\mathrm{F}=345 \mathrm{~N}\). \(\frac{\mathrm{dm}}{\mathrm{dt}}=\) ? \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}}=\) Rate of change in momentum \(\mathrm{F} =\mathrm{v} \frac{\mathrm{dm}}{\mathrm{dt}}\) \(345 =300 \frac{\mathrm{dm}}{\mathrm{dt}}\) \(\frac{\mathrm{dm}}{\mathrm{dt}} =\frac{345}{300}\) \(\frac{\mathrm{dm}}{\mathrm{dt}} =1.15 \mathrm{~kg} / \mathrm{s}\)
AIPMT- 1995
Laws of Motion
146354
A satellite in a force free space sweeps stationary interplanetary dust at a rate. \(\left(\frac{\mathbf{d M}}{\mathbf{d t}}\right)=\alpha \mathrm{v}\). The acceleration of satellite is
1 \(-\frac{2 \alpha v^{2}}{M}\)
2 \(-\frac{\alpha v^{2}}{M}\)
3 \(-\frac{\alpha v^{2}}{2 M}\)
4 \(-\alpha v^{2}\)
Explanation:
B Given, \(\frac{\mathrm{dM}}{\mathrm{dt}}=\alpha \mathrm{v}\) From, Newton's second law of motion \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(F=\frac{d}{d t}(M v)\) \(F=M \frac{d v}{d t}+v \frac{d M}{d t}\) Given the space is force free then net force is zero \((F=\) \(0)\), \(0=\mathrm{M} \frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{v} \frac{\mathrm{dM}}{\mathrm{dt}}\) We know that, \(\quad a=\frac{d v}{d t}\) From equation (i), \(0=\mathrm{Ma}+\mathrm{v}(\alpha \mathrm{v})\) \(-\mathrm{Ma}=\alpha \mathrm{v}^{2}\) \(\mathrm{a}=-\frac{\alpha \cdot \mathrm{v}^{2}}{\mathrm{M}}\)
146350
A rocket with an initial mass \(m_{0}\) is going up with a constant acceleration a by exhausting gases with a velocity \(v\) relative to the rocket motion, then the mass of the rocket at any instant of time is (assume that no other forces act on it)
1 \(m=m_{0} e^{-\frac{a t}{v}}\)
2 \(m=m_{0} e^{-\frac{2 a t}{v}}\)
3 \(m=m_{0} e^{-\frac{a t}{2 v}}\)
4 \(m=m_{0} e^{-\frac{a^{2} t^{2}}{v^{2}}}\)
Explanation:
A Given, External forces on rocket is zero \(\left(\mathrm{F}_{\text {ext }}=0\right)\). External force \(=\) Rate of change in momentum \(F_{\text {ext }}=\frac{d p}{d t}=\frac{d(m v)}{d t}\) \(F_{\text {ext }}=m \frac{d v}{d t}+v \frac{d m}{d t}\) \(m \frac{d v}{d t}+v \frac{d m}{d t}=0\) \(\left(\text { Where } \frac{d v}{d t}=a\right)\) \(\therefore \quad m a=-v \frac{d m}{d t}\) \(\frac{-a}{v} d t=\frac{d m}{m}\) Integrating both side \(\frac{-\mathrm{a}}{\mathrm{v}} \int_{0}^{\mathrm{t}} \mathrm{dt}=\int_{\mathrm{m}_{0}}^{\mathrm{m}} \frac{\mathrm{dm}}{\mathrm{m}}\) \(-\frac{\mathrm{a}}{\mathrm{v}}[\mathrm{t}]_{0}^{\mathrm{t}}=\left[\log _{\mathrm{e}} \mathrm{m}\right]_{\mathrm{m}_{0}}^{\mathrm{m}}\) \(\frac{-\mathrm{at}}{\mathrm{v}}=\log _{\mathrm{e}} \frac{\mathrm{m}}{\mathrm{m}_{0}}\) \(\frac{\mathrm{m}}{\mathrm{m}_{0}}=\mathrm{e}^{-\mathrm{at} / \mathrm{v}}\) \(\mathrm{m}=\mathrm{m}_{0} \mathrm{e}^{-\mathrm{at} / \mathrm{v}}\)
AP EAMCET (23.04.2018) Shift-2
Laws of Motion
146352
A \(600 \mathrm{~kg}\) rocket is set for a vertical firing. If the exhaust speed is \(1000 \mathrm{~ms}^{-1}\), the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is
1 \(117.6 \mathrm{~kg} \mathrm{~s}^{-1}\)
2 \(58.6 \mathrm{~kg} \mathrm{~s}^{-1}\)
3 \(6 \mathrm{~kg} \mathrm{~s}^{-1}\)
4 \(76.4 \mathrm{~kg} \mathrm{~s}^{-1}\)
Explanation:
C Given, \(\mathrm{m}=600 \mathrm{~kg}\) Force required to overcome the weight of rocket is, \(\mathrm{F}=\mathrm{mg}\) \(\frac{\mathrm{dp}}{\mathrm{dt}}=\mathrm{mg}\) \(\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}}=\mathrm{mg}\) \(\mathrm{v} \frac{\mathrm{dm}}{\mathrm{dt}}=\mathrm{mg}\) \(\frac{\mathrm{dm}}{\mathrm{dt}}=\frac{\mathrm{mg}}{\mathrm{v}}\) \(\frac{\mathrm{dm}}{\mathrm{dt}}=\frac{600 \times 10}{1000}\) \(\frac{\mathrm{dm}}{\mathrm{dt}}=6 \mathrm{kgs}^{-1}\) \(\left[\because \frac{\mathrm{dv}}{\mathrm{dt}}=0\right]\)
AIPMT- 1990
Laws of Motion
146353
If the force on a rocket moving with a velocity of \(300 \mathrm{~m} / \mathrm{s}\) is \(345 \mathrm{~N}\), then the rate of combustion of the fuel is
1 \(0.55 \mathrm{~kg} / \mathrm{s}\)
2 \(0.75 \mathrm{~kg} / \mathrm{s}\)
3 \(1.15 \mathrm{~kg} / \mathrm{s}\)
4 \(2.25 \mathrm{~kg} / \mathrm{s}\)
Explanation:
C Given that, \(\mathrm{v}=300 \mathrm{~m} / \mathrm{s}\) \(\mathrm{F}=345 \mathrm{~N}\). \(\frac{\mathrm{dm}}{\mathrm{dt}}=\) ? \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}}=\) Rate of change in momentum \(\mathrm{F} =\mathrm{v} \frac{\mathrm{dm}}{\mathrm{dt}}\) \(345 =300 \frac{\mathrm{dm}}{\mathrm{dt}}\) \(\frac{\mathrm{dm}}{\mathrm{dt}} =\frac{345}{300}\) \(\frac{\mathrm{dm}}{\mathrm{dt}} =1.15 \mathrm{~kg} / \mathrm{s}\)
AIPMT- 1995
Laws of Motion
146354
A satellite in a force free space sweeps stationary interplanetary dust at a rate. \(\left(\frac{\mathbf{d M}}{\mathbf{d t}}\right)=\alpha \mathrm{v}\). The acceleration of satellite is
1 \(-\frac{2 \alpha v^{2}}{M}\)
2 \(-\frac{\alpha v^{2}}{M}\)
3 \(-\frac{\alpha v^{2}}{2 M}\)
4 \(-\alpha v^{2}\)
Explanation:
B Given, \(\frac{\mathrm{dM}}{\mathrm{dt}}=\alpha \mathrm{v}\) From, Newton's second law of motion \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(F=\frac{d}{d t}(M v)\) \(F=M \frac{d v}{d t}+v \frac{d M}{d t}\) Given the space is force free then net force is zero \((F=\) \(0)\), \(0=\mathrm{M} \frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{v} \frac{\mathrm{dM}}{\mathrm{dt}}\) We know that, \(\quad a=\frac{d v}{d t}\) From equation (i), \(0=\mathrm{Ma}+\mathrm{v}(\alpha \mathrm{v})\) \(-\mathrm{Ma}=\alpha \mathrm{v}^{2}\) \(\mathrm{a}=-\frac{\alpha \cdot \mathrm{v}^{2}}{\mathrm{M}}\)
146350
A rocket with an initial mass \(m_{0}\) is going up with a constant acceleration a by exhausting gases with a velocity \(v\) relative to the rocket motion, then the mass of the rocket at any instant of time is (assume that no other forces act on it)
1 \(m=m_{0} e^{-\frac{a t}{v}}\)
2 \(m=m_{0} e^{-\frac{2 a t}{v}}\)
3 \(m=m_{0} e^{-\frac{a t}{2 v}}\)
4 \(m=m_{0} e^{-\frac{a^{2} t^{2}}{v^{2}}}\)
Explanation:
A Given, External forces on rocket is zero \(\left(\mathrm{F}_{\text {ext }}=0\right)\). External force \(=\) Rate of change in momentum \(F_{\text {ext }}=\frac{d p}{d t}=\frac{d(m v)}{d t}\) \(F_{\text {ext }}=m \frac{d v}{d t}+v \frac{d m}{d t}\) \(m \frac{d v}{d t}+v \frac{d m}{d t}=0\) \(\left(\text { Where } \frac{d v}{d t}=a\right)\) \(\therefore \quad m a=-v \frac{d m}{d t}\) \(\frac{-a}{v} d t=\frac{d m}{m}\) Integrating both side \(\frac{-\mathrm{a}}{\mathrm{v}} \int_{0}^{\mathrm{t}} \mathrm{dt}=\int_{\mathrm{m}_{0}}^{\mathrm{m}} \frac{\mathrm{dm}}{\mathrm{m}}\) \(-\frac{\mathrm{a}}{\mathrm{v}}[\mathrm{t}]_{0}^{\mathrm{t}}=\left[\log _{\mathrm{e}} \mathrm{m}\right]_{\mathrm{m}_{0}}^{\mathrm{m}}\) \(\frac{-\mathrm{at}}{\mathrm{v}}=\log _{\mathrm{e}} \frac{\mathrm{m}}{\mathrm{m}_{0}}\) \(\frac{\mathrm{m}}{\mathrm{m}_{0}}=\mathrm{e}^{-\mathrm{at} / \mathrm{v}}\) \(\mathrm{m}=\mathrm{m}_{0} \mathrm{e}^{-\mathrm{at} / \mathrm{v}}\)
AP EAMCET (23.04.2018) Shift-2
Laws of Motion
146352
A \(600 \mathrm{~kg}\) rocket is set for a vertical firing. If the exhaust speed is \(1000 \mathrm{~ms}^{-1}\), the mass of the gas ejected per second to supply the thrust needed to overcome the weight of rocket is
1 \(117.6 \mathrm{~kg} \mathrm{~s}^{-1}\)
2 \(58.6 \mathrm{~kg} \mathrm{~s}^{-1}\)
3 \(6 \mathrm{~kg} \mathrm{~s}^{-1}\)
4 \(76.4 \mathrm{~kg} \mathrm{~s}^{-1}\)
Explanation:
C Given, \(\mathrm{m}=600 \mathrm{~kg}\) Force required to overcome the weight of rocket is, \(\mathrm{F}=\mathrm{mg}\) \(\frac{\mathrm{dp}}{\mathrm{dt}}=\mathrm{mg}\) \(\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}}=\mathrm{mg}\) \(\mathrm{v} \frac{\mathrm{dm}}{\mathrm{dt}}=\mathrm{mg}\) \(\frac{\mathrm{dm}}{\mathrm{dt}}=\frac{\mathrm{mg}}{\mathrm{v}}\) \(\frac{\mathrm{dm}}{\mathrm{dt}}=\frac{600 \times 10}{1000}\) \(\frac{\mathrm{dm}}{\mathrm{dt}}=6 \mathrm{kgs}^{-1}\) \(\left[\because \frac{\mathrm{dv}}{\mathrm{dt}}=0\right]\)
AIPMT- 1990
Laws of Motion
146353
If the force on a rocket moving with a velocity of \(300 \mathrm{~m} / \mathrm{s}\) is \(345 \mathrm{~N}\), then the rate of combustion of the fuel is
1 \(0.55 \mathrm{~kg} / \mathrm{s}\)
2 \(0.75 \mathrm{~kg} / \mathrm{s}\)
3 \(1.15 \mathrm{~kg} / \mathrm{s}\)
4 \(2.25 \mathrm{~kg} / \mathrm{s}\)
Explanation:
C Given that, \(\mathrm{v}=300 \mathrm{~m} / \mathrm{s}\) \(\mathrm{F}=345 \mathrm{~N}\). \(\frac{\mathrm{dm}}{\mathrm{dt}}=\) ? \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}}=\) Rate of change in momentum \(\mathrm{F} =\mathrm{v} \frac{\mathrm{dm}}{\mathrm{dt}}\) \(345 =300 \frac{\mathrm{dm}}{\mathrm{dt}}\) \(\frac{\mathrm{dm}}{\mathrm{dt}} =\frac{345}{300}\) \(\frac{\mathrm{dm}}{\mathrm{dt}} =1.15 \mathrm{~kg} / \mathrm{s}\)
AIPMT- 1995
Laws of Motion
146354
A satellite in a force free space sweeps stationary interplanetary dust at a rate. \(\left(\frac{\mathbf{d M}}{\mathbf{d t}}\right)=\alpha \mathrm{v}\). The acceleration of satellite is
1 \(-\frac{2 \alpha v^{2}}{M}\)
2 \(-\frac{\alpha v^{2}}{M}\)
3 \(-\frac{\alpha v^{2}}{2 M}\)
4 \(-\alpha v^{2}\)
Explanation:
B Given, \(\frac{\mathrm{dM}}{\mathrm{dt}}=\alpha \mathrm{v}\) From, Newton's second law of motion \(\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}\) \(F=\frac{d}{d t}(M v)\) \(F=M \frac{d v}{d t}+v \frac{d M}{d t}\) Given the space is force free then net force is zero \((F=\) \(0)\), \(0=\mathrm{M} \frac{\mathrm{dv}}{\mathrm{dt}}+\mathrm{v} \frac{\mathrm{dM}}{\mathrm{dt}}\) We know that, \(\quad a=\frac{d v}{d t}\) From equation (i), \(0=\mathrm{Ma}+\mathrm{v}(\alpha \mathrm{v})\) \(-\mathrm{Ma}=\alpha \mathrm{v}^{2}\) \(\mathrm{a}=-\frac{\alpha \cdot \mathrm{v}^{2}}{\mathrm{M}}\)
