NEET Test Series from KOTA - 10 Papers In MS WORD
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LAWS OF MOTION (ADDITIONAL)
371744
At a metro station, a girl walks up a stationary escalator in \(20 \mathrm{~s}\). If she remains stationary on the escalator, then the escaltor take her up in 30s. The time taken by her to walk up on the moving escalator will be :
1 \(25 \mathrm{~s}\)
2 \(60 \mathrm{~s}\)
3 \(12 \mathrm{~s}\)
4 \(10 \mathrm{~s}\)
Explanation:
C Given, Let distance \(=\mathrm{h}\) Walking time \(=20 \mathrm{sec}\) Escalator time \(=30 \mathrm{sec}\) Speed of \(\operatorname{girl}\left(\mathrm{v}_{\mathrm{g}}\right)=\frac{\mathrm{h}}{20}\) Speed of escalator \(\left(\mathrm{v}_{\mathrm{e}}\right)=\frac{\mathrm{h}}{30}\) Time taken by girl when she walk upon the moving escalator is \(t=\frac{h}{v_{g}+v_{e}} \Rightarrow t=\frac{h}{\frac{h}{20}+\frac{h}{30}}\) \(=\frac{20 \times 30}{(30+20)}\) \(=12 \mathrm{sec}\)
Karnataka CET-2020
LAWS OF MOTION (ADDITIONAL)
371745
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 \(400 \mathrm{~m} / \mathrm{s}\). Then find out the impulse exerted till force on bullet becomes zero.
371746
The linear momentum of a particle moving in \(X\) - \(Y\) plane under the influence of a force is given as \(\vec{P}(t)=A(\hat{i} \cos b t-\hat{j} \sin b t)\) where \(A\) and \(b\) are constant. The angle between the force and momentum is
1 \(0^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(90^{\circ}\)
Explanation:
D Given, \(\vec{P}=A(\hat{i} \cos b t-\hat{j} \sin b t)\) \(\vec{F}=\frac{d \vec{p}}{d t}=\frac{d}{d t} A(\hat{i} \cos b t-\hat{j} \sin b t)\) \(\vec{F}=A b[-\hat{i} \sin (b t)-\hat{j} \cos (b t)]\) Angle between two vector is \(\theta\) \(\cos \theta=\frac{\mathrm{F} \cdot \mathrm{P}}{|\mathrm{F}| \cdot|\mathrm{P}|}\) \(\cos \theta=\frac{A b(-\hat{i} \sin b t-\hat{j} \cos b t) A(\hat{i} \cos b t-\hat{j} \sin b t)}{|F| \cdot|P|}\) \(\cos \theta=\frac{A^{2} b(-\sin b t \cdot \cos b t+\sin b t \cdot \cos b t)}{|F| \cdot|P|}\) \(\therefore \cos \theta=0\) \(\Rightarrow \cos \theta=\cos 90^{\circ}\) \(\Rightarrow \theta=90^{\circ}\)
UPSEE 2019
LAWS OF MOTION (ADDITIONAL)
371747
Two objects of mass \(m\) each moving with speed u m/s collide at \(90^{\circ}\), then final momentum is (assume collision is inelastic)
1 \(\mathrm{mu}\)
2 \(2 \mathrm{mu}\)
3 \(\sqrt{2} \mathrm{mu}\)
4 \(2 \sqrt{2} \mathrm{mu}\)
Explanation:
C Given, Speed of objects \(=\mathrm{u} \mathrm{m} / \mathrm{s}\) Since, both object collide with \(90^{\circ}\) According to law of conservation of momentum, Total momentum before collision \(=\) Total momentum after collision \(P_{i}=P_{f}\) \(|m u \hat{i}+m u \hat{j}|=P_{f}\) \(\sqrt{m^{2} u^{2}+m^{2} u^{2}}=P_{f}\) \(\qquad P_{f}=\sqrt{2} m u\)
371744
At a metro station, a girl walks up a stationary escalator in \(20 \mathrm{~s}\). If she remains stationary on the escalator, then the escaltor take her up in 30s. The time taken by her to walk up on the moving escalator will be :
1 \(25 \mathrm{~s}\)
2 \(60 \mathrm{~s}\)
3 \(12 \mathrm{~s}\)
4 \(10 \mathrm{~s}\)
Explanation:
C Given, Let distance \(=\mathrm{h}\) Walking time \(=20 \mathrm{sec}\) Escalator time \(=30 \mathrm{sec}\) Speed of \(\operatorname{girl}\left(\mathrm{v}_{\mathrm{g}}\right)=\frac{\mathrm{h}}{20}\) Speed of escalator \(\left(\mathrm{v}_{\mathrm{e}}\right)=\frac{\mathrm{h}}{30}\) Time taken by girl when she walk upon the moving escalator is \(t=\frac{h}{v_{g}+v_{e}} \Rightarrow t=\frac{h}{\frac{h}{20}+\frac{h}{30}}\) \(=\frac{20 \times 30}{(30+20)}\) \(=12 \mathrm{sec}\)
Karnataka CET-2020
LAWS OF MOTION (ADDITIONAL)
371745
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 \(400 \mathrm{~m} / \mathrm{s}\). Then find out the impulse exerted till force on bullet becomes zero.
371746
The linear momentum of a particle moving in \(X\) - \(Y\) plane under the influence of a force is given as \(\vec{P}(t)=A(\hat{i} \cos b t-\hat{j} \sin b t)\) where \(A\) and \(b\) are constant. The angle between the force and momentum is
1 \(0^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(90^{\circ}\)
Explanation:
D Given, \(\vec{P}=A(\hat{i} \cos b t-\hat{j} \sin b t)\) \(\vec{F}=\frac{d \vec{p}}{d t}=\frac{d}{d t} A(\hat{i} \cos b t-\hat{j} \sin b t)\) \(\vec{F}=A b[-\hat{i} \sin (b t)-\hat{j} \cos (b t)]\) Angle between two vector is \(\theta\) \(\cos \theta=\frac{\mathrm{F} \cdot \mathrm{P}}{|\mathrm{F}| \cdot|\mathrm{P}|}\) \(\cos \theta=\frac{A b(-\hat{i} \sin b t-\hat{j} \cos b t) A(\hat{i} \cos b t-\hat{j} \sin b t)}{|F| \cdot|P|}\) \(\cos \theta=\frac{A^{2} b(-\sin b t \cdot \cos b t+\sin b t \cdot \cos b t)}{|F| \cdot|P|}\) \(\therefore \cos \theta=0\) \(\Rightarrow \cos \theta=\cos 90^{\circ}\) \(\Rightarrow \theta=90^{\circ}\)
UPSEE 2019
LAWS OF MOTION (ADDITIONAL)
371747
Two objects of mass \(m\) each moving with speed u m/s collide at \(90^{\circ}\), then final momentum is (assume collision is inelastic)
1 \(\mathrm{mu}\)
2 \(2 \mathrm{mu}\)
3 \(\sqrt{2} \mathrm{mu}\)
4 \(2 \sqrt{2} \mathrm{mu}\)
Explanation:
C Given, Speed of objects \(=\mathrm{u} \mathrm{m} / \mathrm{s}\) Since, both object collide with \(90^{\circ}\) According to law of conservation of momentum, Total momentum before collision \(=\) Total momentum after collision \(P_{i}=P_{f}\) \(|m u \hat{i}+m u \hat{j}|=P_{f}\) \(\sqrt{m^{2} u^{2}+m^{2} u^{2}}=P_{f}\) \(\qquad P_{f}=\sqrt{2} m u\)
371744
At a metro station, a girl walks up a stationary escalator in \(20 \mathrm{~s}\). If she remains stationary on the escalator, then the escaltor take her up in 30s. The time taken by her to walk up on the moving escalator will be :
1 \(25 \mathrm{~s}\)
2 \(60 \mathrm{~s}\)
3 \(12 \mathrm{~s}\)
4 \(10 \mathrm{~s}\)
Explanation:
C Given, Let distance \(=\mathrm{h}\) Walking time \(=20 \mathrm{sec}\) Escalator time \(=30 \mathrm{sec}\) Speed of \(\operatorname{girl}\left(\mathrm{v}_{\mathrm{g}}\right)=\frac{\mathrm{h}}{20}\) Speed of escalator \(\left(\mathrm{v}_{\mathrm{e}}\right)=\frac{\mathrm{h}}{30}\) Time taken by girl when she walk upon the moving escalator is \(t=\frac{h}{v_{g}+v_{e}} \Rightarrow t=\frac{h}{\frac{h}{20}+\frac{h}{30}}\) \(=\frac{20 \times 30}{(30+20)}\) \(=12 \mathrm{sec}\)
Karnataka CET-2020
LAWS OF MOTION (ADDITIONAL)
371745
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 \(400 \mathrm{~m} / \mathrm{s}\). Then find out the impulse exerted till force on bullet becomes zero.
371746
The linear momentum of a particle moving in \(X\) - \(Y\) plane under the influence of a force is given as \(\vec{P}(t)=A(\hat{i} \cos b t-\hat{j} \sin b t)\) where \(A\) and \(b\) are constant. The angle between the force and momentum is
1 \(0^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(90^{\circ}\)
Explanation:
D Given, \(\vec{P}=A(\hat{i} \cos b t-\hat{j} \sin b t)\) \(\vec{F}=\frac{d \vec{p}}{d t}=\frac{d}{d t} A(\hat{i} \cos b t-\hat{j} \sin b t)\) \(\vec{F}=A b[-\hat{i} \sin (b t)-\hat{j} \cos (b t)]\) Angle between two vector is \(\theta\) \(\cos \theta=\frac{\mathrm{F} \cdot \mathrm{P}}{|\mathrm{F}| \cdot|\mathrm{P}|}\) \(\cos \theta=\frac{A b(-\hat{i} \sin b t-\hat{j} \cos b t) A(\hat{i} \cos b t-\hat{j} \sin b t)}{|F| \cdot|P|}\) \(\cos \theta=\frac{A^{2} b(-\sin b t \cdot \cos b t+\sin b t \cdot \cos b t)}{|F| \cdot|P|}\) \(\therefore \cos \theta=0\) \(\Rightarrow \cos \theta=\cos 90^{\circ}\) \(\Rightarrow \theta=90^{\circ}\)
UPSEE 2019
LAWS OF MOTION (ADDITIONAL)
371747
Two objects of mass \(m\) each moving with speed u m/s collide at \(90^{\circ}\), then final momentum is (assume collision is inelastic)
1 \(\mathrm{mu}\)
2 \(2 \mathrm{mu}\)
3 \(\sqrt{2} \mathrm{mu}\)
4 \(2 \sqrt{2} \mathrm{mu}\)
Explanation:
C Given, Speed of objects \(=\mathrm{u} \mathrm{m} / \mathrm{s}\) Since, both object collide with \(90^{\circ}\) According to law of conservation of momentum, Total momentum before collision \(=\) Total momentum after collision \(P_{i}=P_{f}\) \(|m u \hat{i}+m u \hat{j}|=P_{f}\) \(\sqrt{m^{2} u^{2}+m^{2} u^{2}}=P_{f}\) \(\qquad P_{f}=\sqrt{2} m u\)
371744
At a metro station, a girl walks up a stationary escalator in \(20 \mathrm{~s}\). If she remains stationary on the escalator, then the escaltor take her up in 30s. The time taken by her to walk up on the moving escalator will be :
1 \(25 \mathrm{~s}\)
2 \(60 \mathrm{~s}\)
3 \(12 \mathrm{~s}\)
4 \(10 \mathrm{~s}\)
Explanation:
C Given, Let distance \(=\mathrm{h}\) Walking time \(=20 \mathrm{sec}\) Escalator time \(=30 \mathrm{sec}\) Speed of \(\operatorname{girl}\left(\mathrm{v}_{\mathrm{g}}\right)=\frac{\mathrm{h}}{20}\) Speed of escalator \(\left(\mathrm{v}_{\mathrm{e}}\right)=\frac{\mathrm{h}}{30}\) Time taken by girl when she walk upon the moving escalator is \(t=\frac{h}{v_{g}+v_{e}} \Rightarrow t=\frac{h}{\frac{h}{20}+\frac{h}{30}}\) \(=\frac{20 \times 30}{(30+20)}\) \(=12 \mathrm{sec}\)
Karnataka CET-2020
LAWS OF MOTION (ADDITIONAL)
371745
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 \(400 \mathrm{~m} / \mathrm{s}\). Then find out the impulse exerted till force on bullet becomes zero.
371746
The linear momentum of a particle moving in \(X\) - \(Y\) plane under the influence of a force is given as \(\vec{P}(t)=A(\hat{i} \cos b t-\hat{j} \sin b t)\) where \(A\) and \(b\) are constant. The angle between the force and momentum is
1 \(0^{\circ}\)
2 \(45^{\circ}\)
3 \(60^{\circ}\)
4 \(90^{\circ}\)
Explanation:
D Given, \(\vec{P}=A(\hat{i} \cos b t-\hat{j} \sin b t)\) \(\vec{F}=\frac{d \vec{p}}{d t}=\frac{d}{d t} A(\hat{i} \cos b t-\hat{j} \sin b t)\) \(\vec{F}=A b[-\hat{i} \sin (b t)-\hat{j} \cos (b t)]\) Angle between two vector is \(\theta\) \(\cos \theta=\frac{\mathrm{F} \cdot \mathrm{P}}{|\mathrm{F}| \cdot|\mathrm{P}|}\) \(\cos \theta=\frac{A b(-\hat{i} \sin b t-\hat{j} \cos b t) A(\hat{i} \cos b t-\hat{j} \sin b t)}{|F| \cdot|P|}\) \(\cos \theta=\frac{A^{2} b(-\sin b t \cdot \cos b t+\sin b t \cdot \cos b t)}{|F| \cdot|P|}\) \(\therefore \cos \theta=0\) \(\Rightarrow \cos \theta=\cos 90^{\circ}\) \(\Rightarrow \theta=90^{\circ}\)
UPSEE 2019
LAWS OF MOTION (ADDITIONAL)
371747
Two objects of mass \(m\) each moving with speed u m/s collide at \(90^{\circ}\), then final momentum is (assume collision is inelastic)
1 \(\mathrm{mu}\)
2 \(2 \mathrm{mu}\)
3 \(\sqrt{2} \mathrm{mu}\)
4 \(2 \sqrt{2} \mathrm{mu}\)
Explanation:
C Given, Speed of objects \(=\mathrm{u} \mathrm{m} / \mathrm{s}\) Since, both object collide with \(90^{\circ}\) According to law of conservation of momentum, Total momentum before collision \(=\) Total momentum after collision \(P_{i}=P_{f}\) \(|m u \hat{i}+m u \hat{j}|=P_{f}\) \(\sqrt{m^{2} u^{2}+m^{2} u^{2}}=P_{f}\) \(\qquad P_{f}=\sqrt{2} m u\)
