QBManager

Motion in Plane

143737 A swimmer wants to cross a \(200 \mathrm{~m}\) wide river which is flowing at a speed of \(2 \mathrm{~m} / \mathrm{s}\). The velocity of the swimmer with respect to the river is \(1 \mathrm{~m} / \mathrm{s}\). How far from the point directly opposite to the starting point does the swimmer reach the opposite bank?

1 \(200 \mathrm{~m}\)

2 \(400 \mathrm{~m}\)

3 \(600 \mathrm{~m}\)

4 \(800 \mathrm{~m}\)

TSEAMCET(Engg.)-2017

Motion in Plane

143738 At time \(t=0\), a force \(F=\alpha t\), where \(t\) is time in seconds, applied to a body of mass \(1 \mathrm{~kg}\), resting on a smooth horizontal plane. If the direction of the force makes an angle of \(45^{\circ}\) with the horizontal, then the velocity of the body at the moment of its breaking off the plane is

1 \(\frac{100}{\alpha} \mathrm{m} / \mathrm{s}\)

2 \(\frac{50 \sqrt{2}}{\alpha} \mathrm{m} / \mathrm{s}\)

3 \(\frac{50 \alpha}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\)

4 \(\frac{50}{\alpha} \mathrm{m} / \mathrm{s}\)

Explanation:

B Given, \(t=0\), A Force \(F=\alpha t\) is applied to small body of mass \(\mathrm{m}\) resting on a smooth horizontal plane where \(\alpha\) is constant.

The normal reaction will be zero where the body will leave the ground surface.
\(\Rightarrow \mathrm{N}=\mathrm{mg}-\alpha \mathrm{t}_{0} \sin \theta\)
\(\Rightarrow \mathrm{mg}-\alpha \mathrm{t}_{0} \sin \theta=0\)
\(\mathrm{t}_{0}=\frac{\mathrm{mg}}{\alpha \sin \theta}\)
\(\mathrm{t}_{0}=\) time where the body leaves the ground. The force component on horizontal direction is
\(\Rightarrow \mathrm{ma}_{\mathrm{x}}=\alpha \mathrm{t} \cos \theta\)
\(\because \quad \mathrm{a}_{\mathrm{x}}=\frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}\)
\(\therefore \quad \mathrm{m} \times \frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}=\alpha \mathrm{t} \cos \theta\)
\(m \cdot d v_{x}=\alpha \cos \theta \cdot t d t\)
Integrating both side,
\(\mathrm{m} \int_{0}^{\mathrm{v}} d v_{x}=\alpha \cos \theta \int_{0}^{t_{0}} t d t\)
\(m . v=\alpha \cos \theta \cdot\left(\frac{t_{0}{ }^{2}}{2}\right)\)
\(\Rightarrow \quad \mathrm{v}=\frac{\alpha \cos \theta \cdot \mathrm{t}_{0}{ }^{2}}{2 \mathrm{~m}}\)
Put \(\mathrm{t}_{0}=\frac{\mathrm{mg}}{\alpha \sin \theta}\) in equation (iii)
\(\mathrm{v}=\frac{\alpha \cos \theta}{2 \mathrm{~m}} \cdot\left(\frac{\mathrm{mg}}{\alpha \sin \theta}\right)^{2}\)
\(\mathrm{v}=\frac{\alpha \cos \theta}{2 \mathrm{~m}} \cdot \frac{\mathrm{m}^{2} \mathrm{~g}^{2}}{\alpha^{2} \sin ^{2} \theta}\)
\(v=\frac{\alpha \cos \theta \times m^{2} g^{2}}{\alpha^{2} \sin ^{2} \theta 2 m}\)
\(\mathrm{v}=\frac{\mathrm{mg}^{2} \cos \theta}{2 \alpha \sin ^{2} \theta}\)
\(\mathrm{v}=\frac{\mathrm{mg}^{2}}{2 \alpha \sin \theta \cdot \tan \theta}\)
Given, \(\mathrm{m}=1 \mathrm{~kg}, \theta=45^{\circ}\)
\(\mathrm{v}=\frac{1 \times(10)^{2}}{2 \alpha \cdot \sin 45^{\circ} \tan 45^{\circ}} \quad\left\{\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right\}\)
\(\mathrm{v}=\frac{100}{2 \alpha \cdot \frac{1}{\sqrt{2}} \times 1} \Rightarrow \mathrm{v}=\frac{50 \sqrt{2}}{\alpha} \mathrm{m} / \mathrm{s}\)

TS EAMCET 20.07.2022

Motion in Plane

143739 A marble of mass \(m_{1}\) slides down an arc of circular track from rest as shown in the figure. Assume the track is frictionless. If the block having the track has a mass \(\mathrm{m}_{2}\) and can also slide frictionless on the table the velocity of the particle when it exits the track at \(B\) is

1 \(\sqrt{\mathrm{gR}}\)

2 \(\sqrt{g R\left(1+\frac{m_{1}}{m_{2}}\right)}\)

3 \(\sqrt{2 \mathrm{gR}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)}\)

4 \(\sqrt{\frac{2 g \mathrm{gm}_{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}}\)

TS EAMCET 03.05.2018

Motion in Plane

143741 A bus is moving on a straight road towards North with a uniform speed of \(50 \mathrm{~km} / \mathrm{h}\). If the speed remains unchanged after turning through \(90^{\circ}\), the increase in the velocity of bus in the turning process is

1 \(70.7 \mathrm{~km} / \mathrm{h}\) along South-West direction

2 zero

3 \(50 \mathrm{~km} / \mathrm{h}\) along West

4 \(70.7 \mathrm{~km} / \mathrm{h}\) along North-West direction

AIPMT 1989

Motion in Plane

143737 A swimmer wants to cross a \(200 \mathrm{~m}\) wide river which is flowing at a speed of \(2 \mathrm{~m} / \mathrm{s}\). The velocity of the swimmer with respect to the river is \(1 \mathrm{~m} / \mathrm{s}\). How far from the point directly opposite to the starting point does the swimmer reach the opposite bank?

1 \(200 \mathrm{~m}\)

2 \(400 \mathrm{~m}\)

3 \(600 \mathrm{~m}\)

4 \(800 \mathrm{~m}\)

TSEAMCET(Engg.)-2017

Motion in Plane

143738 At time \(t=0\), a force \(F=\alpha t\), where \(t\) is time in seconds, applied to a body of mass \(1 \mathrm{~kg}\), resting on a smooth horizontal plane. If the direction of the force makes an angle of \(45^{\circ}\) with the horizontal, then the velocity of the body at the moment of its breaking off the plane is

1 \(\frac{100}{\alpha} \mathrm{m} / \mathrm{s}\)

2 \(\frac{50 \sqrt{2}}{\alpha} \mathrm{m} / \mathrm{s}\)

3 \(\frac{50 \alpha}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\)

4 \(\frac{50}{\alpha} \mathrm{m} / \mathrm{s}\)

Explanation:

B Given, \(t=0\), A Force \(F=\alpha t\) is applied to small body of mass \(\mathrm{m}\) resting on a smooth horizontal plane where \(\alpha\) is constant.

The normal reaction will be zero where the body will leave the ground surface.
\(\Rightarrow \mathrm{N}=\mathrm{mg}-\alpha \mathrm{t}_{0} \sin \theta\)
\(\Rightarrow \mathrm{mg}-\alpha \mathrm{t}_{0} \sin \theta=0\)
\(\mathrm{t}_{0}=\frac{\mathrm{mg}}{\alpha \sin \theta}\)
\(\mathrm{t}_{0}=\) time where the body leaves the ground. The force component on horizontal direction is
\(\Rightarrow \mathrm{ma}_{\mathrm{x}}=\alpha \mathrm{t} \cos \theta\)
\(\because \quad \mathrm{a}_{\mathrm{x}}=\frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}\)
\(\therefore \quad \mathrm{m} \times \frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}=\alpha \mathrm{t} \cos \theta\)
\(m \cdot d v_{x}=\alpha \cos \theta \cdot t d t\)
Integrating both side,
\(\mathrm{m} \int_{0}^{\mathrm{v}} d v_{x}=\alpha \cos \theta \int_{0}^{t_{0}} t d t\)
\(m . v=\alpha \cos \theta \cdot\left(\frac{t_{0}{ }^{2}}{2}\right)\)
\(\Rightarrow \quad \mathrm{v}=\frac{\alpha \cos \theta \cdot \mathrm{t}_{0}{ }^{2}}{2 \mathrm{~m}}\)
Put \(\mathrm{t}_{0}=\frac{\mathrm{mg}}{\alpha \sin \theta}\) in equation (iii)
\(\mathrm{v}=\frac{\alpha \cos \theta}{2 \mathrm{~m}} \cdot\left(\frac{\mathrm{mg}}{\alpha \sin \theta}\right)^{2}\)
\(\mathrm{v}=\frac{\alpha \cos \theta}{2 \mathrm{~m}} \cdot \frac{\mathrm{m}^{2} \mathrm{~g}^{2}}{\alpha^{2} \sin ^{2} \theta}\)
\(v=\frac{\alpha \cos \theta \times m^{2} g^{2}}{\alpha^{2} \sin ^{2} \theta 2 m}\)
\(\mathrm{v}=\frac{\mathrm{mg}^{2} \cos \theta}{2 \alpha \sin ^{2} \theta}\)
\(\mathrm{v}=\frac{\mathrm{mg}^{2}}{2 \alpha \sin \theta \cdot \tan \theta}\)
Given, \(\mathrm{m}=1 \mathrm{~kg}, \theta=45^{\circ}\)
\(\mathrm{v}=\frac{1 \times(10)^{2}}{2 \alpha \cdot \sin 45^{\circ} \tan 45^{\circ}} \quad\left\{\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right\}\)
\(\mathrm{v}=\frac{100}{2 \alpha \cdot \frac{1}{\sqrt{2}} \times 1} \Rightarrow \mathrm{v}=\frac{50 \sqrt{2}}{\alpha} \mathrm{m} / \mathrm{s}\)

TS EAMCET 20.07.2022

Motion in Plane

143739 A marble of mass \(m_{1}\) slides down an arc of circular track from rest as shown in the figure. Assume the track is frictionless. If the block having the track has a mass \(\mathrm{m}_{2}\) and can also slide frictionless on the table the velocity of the particle when it exits the track at \(B\) is

1 \(\sqrt{\mathrm{gR}}\)

2 \(\sqrt{g R\left(1+\frac{m_{1}}{m_{2}}\right)}\)

3 \(\sqrt{2 \mathrm{gR}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)}\)

4 \(\sqrt{\frac{2 g \mathrm{gm}_{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}}\)

TS EAMCET 03.05.2018

Motion in Plane

143741 A bus is moving on a straight road towards North with a uniform speed of \(50 \mathrm{~km} / \mathrm{h}\). If the speed remains unchanged after turning through \(90^{\circ}\), the increase in the velocity of bus in the turning process is

1 \(70.7 \mathrm{~km} / \mathrm{h}\) along South-West direction

2 zero

3 \(50 \mathrm{~km} / \mathrm{h}\) along West

4 \(70.7 \mathrm{~km} / \mathrm{h}\) along North-West direction

AIPMT 1989

Motion in Plane

143737 A swimmer wants to cross a \(200 \mathrm{~m}\) wide river which is flowing at a speed of \(2 \mathrm{~m} / \mathrm{s}\). The velocity of the swimmer with respect to the river is \(1 \mathrm{~m} / \mathrm{s}\). How far from the point directly opposite to the starting point does the swimmer reach the opposite bank?

1 \(200 \mathrm{~m}\)

2 \(400 \mathrm{~m}\)

3 \(600 \mathrm{~m}\)

4 \(800 \mathrm{~m}\)

TSEAMCET(Engg.)-2017

Motion in Plane

143738 At time \(t=0\), a force \(F=\alpha t\), where \(t\) is time in seconds, applied to a body of mass \(1 \mathrm{~kg}\), resting on a smooth horizontal plane. If the direction of the force makes an angle of \(45^{\circ}\) with the horizontal, then the velocity of the body at the moment of its breaking off the plane is

1 \(\frac{100}{\alpha} \mathrm{m} / \mathrm{s}\)

2 \(\frac{50 \sqrt{2}}{\alpha} \mathrm{m} / \mathrm{s}\)

3 \(\frac{50 \alpha}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\)

4 \(\frac{50}{\alpha} \mathrm{m} / \mathrm{s}\)

Explanation:

B Given, \(t=0\), A Force \(F=\alpha t\) is applied to small body of mass \(\mathrm{m}\) resting on a smooth horizontal plane where \(\alpha\) is constant.

The normal reaction will be zero where the body will leave the ground surface.
\(\Rightarrow \mathrm{N}=\mathrm{mg}-\alpha \mathrm{t}_{0} \sin \theta\)
\(\Rightarrow \mathrm{mg}-\alpha \mathrm{t}_{0} \sin \theta=0\)
\(\mathrm{t}_{0}=\frac{\mathrm{mg}}{\alpha \sin \theta}\)
\(\mathrm{t}_{0}=\) time where the body leaves the ground. The force component on horizontal direction is
\(\Rightarrow \mathrm{ma}_{\mathrm{x}}=\alpha \mathrm{t} \cos \theta\)
\(\because \quad \mathrm{a}_{\mathrm{x}}=\frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}\)
\(\therefore \quad \mathrm{m} \times \frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}=\alpha \mathrm{t} \cos \theta\)
\(m \cdot d v_{x}=\alpha \cos \theta \cdot t d t\)
Integrating both side,
\(\mathrm{m} \int_{0}^{\mathrm{v}} d v_{x}=\alpha \cos \theta \int_{0}^{t_{0}} t d t\)
\(m . v=\alpha \cos \theta \cdot\left(\frac{t_{0}{ }^{2}}{2}\right)\)
\(\Rightarrow \quad \mathrm{v}=\frac{\alpha \cos \theta \cdot \mathrm{t}_{0}{ }^{2}}{2 \mathrm{~m}}\)
Put \(\mathrm{t}_{0}=\frac{\mathrm{mg}}{\alpha \sin \theta}\) in equation (iii)
\(\mathrm{v}=\frac{\alpha \cos \theta}{2 \mathrm{~m}} \cdot\left(\frac{\mathrm{mg}}{\alpha \sin \theta}\right)^{2}\)
\(\mathrm{v}=\frac{\alpha \cos \theta}{2 \mathrm{~m}} \cdot \frac{\mathrm{m}^{2} \mathrm{~g}^{2}}{\alpha^{2} \sin ^{2} \theta}\)
\(v=\frac{\alpha \cos \theta \times m^{2} g^{2}}{\alpha^{2} \sin ^{2} \theta 2 m}\)
\(\mathrm{v}=\frac{\mathrm{mg}^{2} \cos \theta}{2 \alpha \sin ^{2} \theta}\)
\(\mathrm{v}=\frac{\mathrm{mg}^{2}}{2 \alpha \sin \theta \cdot \tan \theta}\)
Given, \(\mathrm{m}=1 \mathrm{~kg}, \theta=45^{\circ}\)
\(\mathrm{v}=\frac{1 \times(10)^{2}}{2 \alpha \cdot \sin 45^{\circ} \tan 45^{\circ}} \quad\left\{\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right\}\)
\(\mathrm{v}=\frac{100}{2 \alpha \cdot \frac{1}{\sqrt{2}} \times 1} \Rightarrow \mathrm{v}=\frac{50 \sqrt{2}}{\alpha} \mathrm{m} / \mathrm{s}\)

TS EAMCET 20.07.2022

Motion in Plane

143739 A marble of mass \(m_{1}\) slides down an arc of circular track from rest as shown in the figure. Assume the track is frictionless. If the block having the track has a mass \(\mathrm{m}_{2}\) and can also slide frictionless on the table the velocity of the particle when it exits the track at \(B\) is

1 \(\sqrt{\mathrm{gR}}\)

2 \(\sqrt{g R\left(1+\frac{m_{1}}{m_{2}}\right)}\)

3 \(\sqrt{2 \mathrm{gR}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)}\)

4 \(\sqrt{\frac{2 g \mathrm{gm}_{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}}\)

TS EAMCET 03.05.2018

Motion in Plane

143741 A bus is moving on a straight road towards North with a uniform speed of \(50 \mathrm{~km} / \mathrm{h}\). If the speed remains unchanged after turning through \(90^{\circ}\), the increase in the velocity of bus in the turning process is

1 \(70.7 \mathrm{~km} / \mathrm{h}\) along South-West direction

2 zero

3 \(50 \mathrm{~km} / \mathrm{h}\) along West

4 \(70.7 \mathrm{~km} / \mathrm{h}\) along North-West direction

AIPMT 1989

Motion in Plane

143737 A swimmer wants to cross a \(200 \mathrm{~m}\) wide river which is flowing at a speed of \(2 \mathrm{~m} / \mathrm{s}\). The velocity of the swimmer with respect to the river is \(1 \mathrm{~m} / \mathrm{s}\). How far from the point directly opposite to the starting point does the swimmer reach the opposite bank?

1 \(200 \mathrm{~m}\)

2 \(400 \mathrm{~m}\)

3 \(600 \mathrm{~m}\)

4 \(800 \mathrm{~m}\)

TSEAMCET(Engg.)-2017

Motion in Plane

143738 At time \(t=0\), a force \(F=\alpha t\), where \(t\) is time in seconds, applied to a body of mass \(1 \mathrm{~kg}\), resting on a smooth horizontal plane. If the direction of the force makes an angle of \(45^{\circ}\) with the horizontal, then the velocity of the body at the moment of its breaking off the plane is

1 \(\frac{100}{\alpha} \mathrm{m} / \mathrm{s}\)

2 \(\frac{50 \sqrt{2}}{\alpha} \mathrm{m} / \mathrm{s}\)

3 \(\frac{50 \alpha}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\)

4 \(\frac{50}{\alpha} \mathrm{m} / \mathrm{s}\)

Explanation:

B Given, \(t=0\), A Force \(F=\alpha t\) is applied to small body of mass \(\mathrm{m}\) resting on a smooth horizontal plane where \(\alpha\) is constant.

The normal reaction will be zero where the body will leave the ground surface.
\(\Rightarrow \mathrm{N}=\mathrm{mg}-\alpha \mathrm{t}_{0} \sin \theta\)
\(\Rightarrow \mathrm{mg}-\alpha \mathrm{t}_{0} \sin \theta=0\)
\(\mathrm{t}_{0}=\frac{\mathrm{mg}}{\alpha \sin \theta}\)
\(\mathrm{t}_{0}=\) time where the body leaves the ground. The force component on horizontal direction is
\(\Rightarrow \mathrm{ma}_{\mathrm{x}}=\alpha \mathrm{t} \cos \theta\)
\(\because \quad \mathrm{a}_{\mathrm{x}}=\frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}\)
\(\therefore \quad \mathrm{m} \times \frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}=\alpha \mathrm{t} \cos \theta\)
\(m \cdot d v_{x}=\alpha \cos \theta \cdot t d t\)
Integrating both side,
\(\mathrm{m} \int_{0}^{\mathrm{v}} d v_{x}=\alpha \cos \theta \int_{0}^{t_{0}} t d t\)
\(m . v=\alpha \cos \theta \cdot\left(\frac{t_{0}{ }^{2}}{2}\right)\)
\(\Rightarrow \quad \mathrm{v}=\frac{\alpha \cos \theta \cdot \mathrm{t}_{0}{ }^{2}}{2 \mathrm{~m}}\)
Put \(\mathrm{t}_{0}=\frac{\mathrm{mg}}{\alpha \sin \theta}\) in equation (iii)
\(\mathrm{v}=\frac{\alpha \cos \theta}{2 \mathrm{~m}} \cdot\left(\frac{\mathrm{mg}}{\alpha \sin \theta}\right)^{2}\)
\(\mathrm{v}=\frac{\alpha \cos \theta}{2 \mathrm{~m}} \cdot \frac{\mathrm{m}^{2} \mathrm{~g}^{2}}{\alpha^{2} \sin ^{2} \theta}\)
\(v=\frac{\alpha \cos \theta \times m^{2} g^{2}}{\alpha^{2} \sin ^{2} \theta 2 m}\)
\(\mathrm{v}=\frac{\mathrm{mg}^{2} \cos \theta}{2 \alpha \sin ^{2} \theta}\)
\(\mathrm{v}=\frac{\mathrm{mg}^{2}}{2 \alpha \sin \theta \cdot \tan \theta}\)
Given, \(\mathrm{m}=1 \mathrm{~kg}, \theta=45^{\circ}\)
\(\mathrm{v}=\frac{1 \times(10)^{2}}{2 \alpha \cdot \sin 45^{\circ} \tan 45^{\circ}} \quad\left\{\because \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right\}\)
\(\mathrm{v}=\frac{100}{2 \alpha \cdot \frac{1}{\sqrt{2}} \times 1} \Rightarrow \mathrm{v}=\frac{50 \sqrt{2}}{\alpha} \mathrm{m} / \mathrm{s}\)

TS EAMCET 20.07.2022

Motion in Plane

143739 A marble of mass \(m_{1}\) slides down an arc of circular track from rest as shown in the figure. Assume the track is frictionless. If the block having the track has a mass \(\mathrm{m}_{2}\) and can also slide frictionless on the table the velocity of the particle when it exits the track at \(B\) is

1 \(\sqrt{\mathrm{gR}}\)

2 \(\sqrt{g R\left(1+\frac{m_{1}}{m_{2}}\right)}\)

3 \(\sqrt{2 \mathrm{gR}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)}\)

4 \(\sqrt{\frac{2 g \mathrm{gm}_{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}}\)

TS EAMCET 03.05.2018

Motion in Plane

143741 A bus is moving on a straight road towards North with a uniform speed of \(50 \mathrm{~km} / \mathrm{h}\). If the speed remains unchanged after turning through \(90^{\circ}\), the increase in the velocity of bus in the turning process is

1 \(70.7 \mathrm{~km} / \mathrm{h}\) along South-West direction

2 zero

3 \(50 \mathrm{~km} / \mathrm{h}\) along West

4 \(70.7 \mathrm{~km} / \mathrm{h}\) along North-West direction

AIPMT 1989