363502
A gun applies a force \(F\) on a bullet which is given by \(F=\left(100-0.5 \times 10^{5} t\right) N\). The bullet emerges out with speed \(400\;m{\rm{/}}s\). Then, find out the impulse exerted till force on bullet becomes zero.
1 \(0.2\;N{\rm{ - }}s\)
2 \(0.3\;N{\rm{ - }}s\)
3 \(0.1\;N{\rm{ - }}s\)
4 \(0.4\;N{\rm{ - }}s\)
Explanation:
Force applied by gun, \(F=\left(100-0.5 \times 10^{5} t\right) N\) Speed of bullet, \(v = 400\;m{\rm{/}}s\), When \(F=0\), then \(100-0.5 \times 10^{5} t=0\) \(\Rightarrow t=2 \times 10^{-3} s\) Impulse, \(I=\int F d t\) \(=\int_{0}^{2 \times 10^{-3}}\left(100-0.5 \times 10^{5} t\right) d t\) \(=\left[100 t-0.5 \times 10^{5} \cdot \dfrac{t^{2}}{2}\right]^{2 \times 10^{-3}}\) \(=100 \times 2 \times 10^{-3}-\dfrac{0.5 \times 10^{5}}{2} \times 4 \times 10^{-6}-0\) \(=0.2-0.1=0.1 N-s\)
AIIMS - 2019
PHXI05:LAWS OF MOTION
363503
A particle moves in the \(x\)-\(y\) plane under the influence of a force such that its linear momentum is \(\vec p\,(t) = A\,[\hat i\cos (kt) - \hat j\sin (kt)]\) where \(A\) and \(k\) are constants. The angle between the force and momentum is
1 \(0^\circ \)
2 \(30^\circ \)
3 \(45^\circ \)
4 \(90^\circ \)
Explanation:
Given that \(\vec p\,(t) = A(\hat i\cos kt - \hat j\sin kt)\) \(\vec F = \frac{d}{{dt}}(\vec p(t)) = Ak( - \hat i\sin kt - \hat j\cos kt)\) \(\vec F \cdot \vec p = {A^2}k( - \cos kt\sin kt + \sin kt\cos kt) = 0\) \(\therefore \) The momentum and force are perpendicular to each other at \(90^\circ \).
PHXI05:LAWS OF MOTION
363504
The position-time graph for two cars of the same mass is given. The ratio of momentum of the \({\operatorname{car} A}\) and \({\operatorname{car} B}\) is
1 \({3: 1}\)
2 \({1: 3}\)
3 \({1: \sqrt{2}}\)
4 \({\sqrt{3}: 2}\)
Explanation:
Angle of the line \({A}\) is \({30^{\circ}}\) and that of \({B}\) is \({60^{\circ}}\) \(\begin{aligned}& v_{B}=\tan 60^{\circ}=\sqrt{3} \\& v_{A}=\tan 30^{\circ}=\dfrac{1}{\sqrt{3}} \\& \dfrac{v_{A}}{v_{B}}=\dfrac{1}{3}\end{aligned}\)
PHXI05:LAWS OF MOTION
363505
The area of \(F\)-\(t\) curve is \(A\), where ‘\(F\)’ is the force acting on one mass due to the other. If one of the colliding bodies of mass \(M\) is at rest initially, its speed just after the collision is:
1 \(\sqrt {\frac{{2\,A}}{M}} \)
2 \(M/A\)
3 \(AM\)
4 \(A/M\)
Explanation:
The charge in momentum is equal to area of the \(F - t\) curve. Here the mass is at rest initially \(Mv = A \Rightarrow v = \frac{A}{M}\)
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PHXI05:LAWS OF MOTION
363502
A gun applies a force \(F\) on a bullet which is given by \(F=\left(100-0.5 \times 10^{5} t\right) N\). The bullet emerges out with speed \(400\;m{\rm{/}}s\). Then, find out the impulse exerted till force on bullet becomes zero.
1 \(0.2\;N{\rm{ - }}s\)
2 \(0.3\;N{\rm{ - }}s\)
3 \(0.1\;N{\rm{ - }}s\)
4 \(0.4\;N{\rm{ - }}s\)
Explanation:
Force applied by gun, \(F=\left(100-0.5 \times 10^{5} t\right) N\) Speed of bullet, \(v = 400\;m{\rm{/}}s\), When \(F=0\), then \(100-0.5 \times 10^{5} t=0\) \(\Rightarrow t=2 \times 10^{-3} s\) Impulse, \(I=\int F d t\) \(=\int_{0}^{2 \times 10^{-3}}\left(100-0.5 \times 10^{5} t\right) d t\) \(=\left[100 t-0.5 \times 10^{5} \cdot \dfrac{t^{2}}{2}\right]^{2 \times 10^{-3}}\) \(=100 \times 2 \times 10^{-3}-\dfrac{0.5 \times 10^{5}}{2} \times 4 \times 10^{-6}-0\) \(=0.2-0.1=0.1 N-s\)
AIIMS - 2019
PHXI05:LAWS OF MOTION
363503
A particle moves in the \(x\)-\(y\) plane under the influence of a force such that its linear momentum is \(\vec p\,(t) = A\,[\hat i\cos (kt) - \hat j\sin (kt)]\) where \(A\) and \(k\) are constants. The angle between the force and momentum is
1 \(0^\circ \)
2 \(30^\circ \)
3 \(45^\circ \)
4 \(90^\circ \)
Explanation:
Given that \(\vec p\,(t) = A(\hat i\cos kt - \hat j\sin kt)\) \(\vec F = \frac{d}{{dt}}(\vec p(t)) = Ak( - \hat i\sin kt - \hat j\cos kt)\) \(\vec F \cdot \vec p = {A^2}k( - \cos kt\sin kt + \sin kt\cos kt) = 0\) \(\therefore \) The momentum and force are perpendicular to each other at \(90^\circ \).
PHXI05:LAWS OF MOTION
363504
The position-time graph for two cars of the same mass is given. The ratio of momentum of the \({\operatorname{car} A}\) and \({\operatorname{car} B}\) is
1 \({3: 1}\)
2 \({1: 3}\)
3 \({1: \sqrt{2}}\)
4 \({\sqrt{3}: 2}\)
Explanation:
Angle of the line \({A}\) is \({30^{\circ}}\) and that of \({B}\) is \({60^{\circ}}\) \(\begin{aligned}& v_{B}=\tan 60^{\circ}=\sqrt{3} \\& v_{A}=\tan 30^{\circ}=\dfrac{1}{\sqrt{3}} \\& \dfrac{v_{A}}{v_{B}}=\dfrac{1}{3}\end{aligned}\)
PHXI05:LAWS OF MOTION
363505
The area of \(F\)-\(t\) curve is \(A\), where ‘\(F\)’ is the force acting on one mass due to the other. If one of the colliding bodies of mass \(M\) is at rest initially, its speed just after the collision is:
1 \(\sqrt {\frac{{2\,A}}{M}} \)
2 \(M/A\)
3 \(AM\)
4 \(A/M\)
Explanation:
The charge in momentum is equal to area of the \(F - t\) curve. Here the mass is at rest initially \(Mv = A \Rightarrow v = \frac{A}{M}\)
363502
A gun applies a force \(F\) on a bullet which is given by \(F=\left(100-0.5 \times 10^{5} t\right) N\). The bullet emerges out with speed \(400\;m{\rm{/}}s\). Then, find out the impulse exerted till force on bullet becomes zero.
1 \(0.2\;N{\rm{ - }}s\)
2 \(0.3\;N{\rm{ - }}s\)
3 \(0.1\;N{\rm{ - }}s\)
4 \(0.4\;N{\rm{ - }}s\)
Explanation:
Force applied by gun, \(F=\left(100-0.5 \times 10^{5} t\right) N\) Speed of bullet, \(v = 400\;m{\rm{/}}s\), When \(F=0\), then \(100-0.5 \times 10^{5} t=0\) \(\Rightarrow t=2 \times 10^{-3} s\) Impulse, \(I=\int F d t\) \(=\int_{0}^{2 \times 10^{-3}}\left(100-0.5 \times 10^{5} t\right) d t\) \(=\left[100 t-0.5 \times 10^{5} \cdot \dfrac{t^{2}}{2}\right]^{2 \times 10^{-3}}\) \(=100 \times 2 \times 10^{-3}-\dfrac{0.5 \times 10^{5}}{2} \times 4 \times 10^{-6}-0\) \(=0.2-0.1=0.1 N-s\)
AIIMS - 2019
PHXI05:LAWS OF MOTION
363503
A particle moves in the \(x\)-\(y\) plane under the influence of a force such that its linear momentum is \(\vec p\,(t) = A\,[\hat i\cos (kt) - \hat j\sin (kt)]\) where \(A\) and \(k\) are constants. The angle between the force and momentum is
1 \(0^\circ \)
2 \(30^\circ \)
3 \(45^\circ \)
4 \(90^\circ \)
Explanation:
Given that \(\vec p\,(t) = A(\hat i\cos kt - \hat j\sin kt)\) \(\vec F = \frac{d}{{dt}}(\vec p(t)) = Ak( - \hat i\sin kt - \hat j\cos kt)\) \(\vec F \cdot \vec p = {A^2}k( - \cos kt\sin kt + \sin kt\cos kt) = 0\) \(\therefore \) The momentum and force are perpendicular to each other at \(90^\circ \).
PHXI05:LAWS OF MOTION
363504
The position-time graph for two cars of the same mass is given. The ratio of momentum of the \({\operatorname{car} A}\) and \({\operatorname{car} B}\) is
1 \({3: 1}\)
2 \({1: 3}\)
3 \({1: \sqrt{2}}\)
4 \({\sqrt{3}: 2}\)
Explanation:
Angle of the line \({A}\) is \({30^{\circ}}\) and that of \({B}\) is \({60^{\circ}}\) \(\begin{aligned}& v_{B}=\tan 60^{\circ}=\sqrt{3} \\& v_{A}=\tan 30^{\circ}=\dfrac{1}{\sqrt{3}} \\& \dfrac{v_{A}}{v_{B}}=\dfrac{1}{3}\end{aligned}\)
PHXI05:LAWS OF MOTION
363505
The area of \(F\)-\(t\) curve is \(A\), where ‘\(F\)’ is the force acting on one mass due to the other. If one of the colliding bodies of mass \(M\) is at rest initially, its speed just after the collision is:
1 \(\sqrt {\frac{{2\,A}}{M}} \)
2 \(M/A\)
3 \(AM\)
4 \(A/M\)
Explanation:
The charge in momentum is equal to area of the \(F - t\) curve. Here the mass is at rest initially \(Mv = A \Rightarrow v = \frac{A}{M}\)
363502
A gun applies a force \(F\) on a bullet which is given by \(F=\left(100-0.5 \times 10^{5} t\right) N\). The bullet emerges out with speed \(400\;m{\rm{/}}s\). Then, find out the impulse exerted till force on bullet becomes zero.
1 \(0.2\;N{\rm{ - }}s\)
2 \(0.3\;N{\rm{ - }}s\)
3 \(0.1\;N{\rm{ - }}s\)
4 \(0.4\;N{\rm{ - }}s\)
Explanation:
Force applied by gun, \(F=\left(100-0.5 \times 10^{5} t\right) N\) Speed of bullet, \(v = 400\;m{\rm{/}}s\), When \(F=0\), then \(100-0.5 \times 10^{5} t=0\) \(\Rightarrow t=2 \times 10^{-3} s\) Impulse, \(I=\int F d t\) \(=\int_{0}^{2 \times 10^{-3}}\left(100-0.5 \times 10^{5} t\right) d t\) \(=\left[100 t-0.5 \times 10^{5} \cdot \dfrac{t^{2}}{2}\right]^{2 \times 10^{-3}}\) \(=100 \times 2 \times 10^{-3}-\dfrac{0.5 \times 10^{5}}{2} \times 4 \times 10^{-6}-0\) \(=0.2-0.1=0.1 N-s\)
AIIMS - 2019
PHXI05:LAWS OF MOTION
363503
A particle moves in the \(x\)-\(y\) plane under the influence of a force such that its linear momentum is \(\vec p\,(t) = A\,[\hat i\cos (kt) - \hat j\sin (kt)]\) where \(A\) and \(k\) are constants. The angle between the force and momentum is
1 \(0^\circ \)
2 \(30^\circ \)
3 \(45^\circ \)
4 \(90^\circ \)
Explanation:
Given that \(\vec p\,(t) = A(\hat i\cos kt - \hat j\sin kt)\) \(\vec F = \frac{d}{{dt}}(\vec p(t)) = Ak( - \hat i\sin kt - \hat j\cos kt)\) \(\vec F \cdot \vec p = {A^2}k( - \cos kt\sin kt + \sin kt\cos kt) = 0\) \(\therefore \) The momentum and force are perpendicular to each other at \(90^\circ \).
PHXI05:LAWS OF MOTION
363504
The position-time graph for two cars of the same mass is given. The ratio of momentum of the \({\operatorname{car} A}\) and \({\operatorname{car} B}\) is
1 \({3: 1}\)
2 \({1: 3}\)
3 \({1: \sqrt{2}}\)
4 \({\sqrt{3}: 2}\)
Explanation:
Angle of the line \({A}\) is \({30^{\circ}}\) and that of \({B}\) is \({60^{\circ}}\) \(\begin{aligned}& v_{B}=\tan 60^{\circ}=\sqrt{3} \\& v_{A}=\tan 30^{\circ}=\dfrac{1}{\sqrt{3}} \\& \dfrac{v_{A}}{v_{B}}=\dfrac{1}{3}\end{aligned}\)
PHXI05:LAWS OF MOTION
363505
The area of \(F\)-\(t\) curve is \(A\), where ‘\(F\)’ is the force acting on one mass due to the other. If one of the colliding bodies of mass \(M\) is at rest initially, its speed just after the collision is:
1 \(\sqrt {\frac{{2\,A}}{M}} \)
2 \(M/A\)
3 \(AM\)
4 \(A/M\)
Explanation:
The charge in momentum is equal to area of the \(F - t\) curve. Here the mass is at rest initially \(Mv = A \Rightarrow v = \frac{A}{M}\)
