148948
A stationary object breaks into two pieces $A$ and $B$ of masses $6 \mathrm{~kg}$ and $8 \mathrm{~kg}$ respectively. If the velocity of $B$ is $6 \mathrm{~ms}^{-1}$, then find the kinetic energy of $A$.
1 $48 \mathrm{~J}$
2 $192 \mathrm{~J}$
3 $24 \mathrm{~J}$
4 $288 \mathrm{~J}$
Explanation:
B Given, $\mathrm{m}_{\mathrm{A}}=6 \mathrm{~kg}$ $\mathrm{~m}_{\mathrm{B}}=8 \mathrm{~kg}$ $\mathrm{v}_{\mathrm{B}}=6 \mathrm{~m} / \mathrm{sec}$ By the law of conservation of momentum, $\mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}$ $6 \times \mathrm{v}_{\mathrm{A}}=8 \times 6$ $\mathrm{v}_{\mathrm{A}}=8 \mathrm{~m} / \mathrm{sec}$ Kinetic energy of body A $(\mathrm{K} . \mathrm{E})_{\mathrm{A}} =\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}^{2}$ $=\frac{1}{2} \times 6 \times(8)^{2}$ $=\frac{1}{2} \times 6 \times 64$ $\mathrm{KE}_{\mathrm{A}}= 192 \mathrm{~J}$ The Kinetic Energy of body A is $192 \mathrm{~J}$.
AP EAMCET-06.09.2021
Work, Energy and Power
148949
Two bodies having kinetic energy in the ratio 4 : 1 are moving with same linear velocity. The ratio of their masses is
1 $1: 2$
2 $1: 1$
3 $4: 1$
4 $1: 4$
Explanation:
C Given, $\frac{\mathrm{K} . \mathrm{E}_{1}}{\mathrm{~K} . \mathrm{E}_{2}}=\frac{4}{1}$ We know, $\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{~K} \cdot \mathrm{E} \propto \mathrm{mv}^{2}$ $\frac{\mathrm{K} \cdot \mathrm{E}_{1}}{\mathrm{~K} \cdot \mathrm{E}_{2}}=\frac{\mathrm{m}_{1} \mathrm{v}_{1}^{2}}{\mathrm{~m}_{2} \mathrm{v}_{2}^{2}}$ Given, same linear velocity - $\mathrm{v}_{1}=\mathrm{v}_{2}$ $\frac{4}{1}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}$ The ratio $\mathrm{m}_{1}: \mathrm{m}_{2}=4: 1$
AP EAMCET-20.08.2021
Work, Energy and Power
148950
What is the shape of the graph between speed and kinetic energy of a body?
1 A straight line
2 A hyperbola
3 A parabola
4 Exponential
Explanation:
C Let $\mathrm{X}$-axis denotes velocity $\mathrm{v}$ and $\mathrm{Y}$-axis denotes kinetic energy K.E. Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{KE} \propto \mathrm{v}^{2}$ $\mathrm{K} . \mathrm{E}=\mathrm{cv}^{2}$ Where $\mathrm{c}=\frac{1}{2} \mathrm{~m}$ is constant Convert equation (i) in $\mathrm{X}-\mathrm{Y}$ form, $\mathrm{Y}=\mathrm{c} \mathrm{X}^{2}$ So, the graph between speed and kinetic energy of body is parabolic in shape.
AP EAMCET-19.08.2021
Work, Energy and Power
148951
A particle slides down along an inclined plane AC. If the plane is frictionless, kinetic energy of particle at ' $C$ ' is
148948
A stationary object breaks into two pieces $A$ and $B$ of masses $6 \mathrm{~kg}$ and $8 \mathrm{~kg}$ respectively. If the velocity of $B$ is $6 \mathrm{~ms}^{-1}$, then find the kinetic energy of $A$.
1 $48 \mathrm{~J}$
2 $192 \mathrm{~J}$
3 $24 \mathrm{~J}$
4 $288 \mathrm{~J}$
Explanation:
B Given, $\mathrm{m}_{\mathrm{A}}=6 \mathrm{~kg}$ $\mathrm{~m}_{\mathrm{B}}=8 \mathrm{~kg}$ $\mathrm{v}_{\mathrm{B}}=6 \mathrm{~m} / \mathrm{sec}$ By the law of conservation of momentum, $\mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}$ $6 \times \mathrm{v}_{\mathrm{A}}=8 \times 6$ $\mathrm{v}_{\mathrm{A}}=8 \mathrm{~m} / \mathrm{sec}$ Kinetic energy of body A $(\mathrm{K} . \mathrm{E})_{\mathrm{A}} =\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}^{2}$ $=\frac{1}{2} \times 6 \times(8)^{2}$ $=\frac{1}{2} \times 6 \times 64$ $\mathrm{KE}_{\mathrm{A}}= 192 \mathrm{~J}$ The Kinetic Energy of body A is $192 \mathrm{~J}$.
AP EAMCET-06.09.2021
Work, Energy and Power
148949
Two bodies having kinetic energy in the ratio 4 : 1 are moving with same linear velocity. The ratio of their masses is
1 $1: 2$
2 $1: 1$
3 $4: 1$
4 $1: 4$
Explanation:
C Given, $\frac{\mathrm{K} . \mathrm{E}_{1}}{\mathrm{~K} . \mathrm{E}_{2}}=\frac{4}{1}$ We know, $\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{~K} \cdot \mathrm{E} \propto \mathrm{mv}^{2}$ $\frac{\mathrm{K} \cdot \mathrm{E}_{1}}{\mathrm{~K} \cdot \mathrm{E}_{2}}=\frac{\mathrm{m}_{1} \mathrm{v}_{1}^{2}}{\mathrm{~m}_{2} \mathrm{v}_{2}^{2}}$ Given, same linear velocity - $\mathrm{v}_{1}=\mathrm{v}_{2}$ $\frac{4}{1}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}$ The ratio $\mathrm{m}_{1}: \mathrm{m}_{2}=4: 1$
AP EAMCET-20.08.2021
Work, Energy and Power
148950
What is the shape of the graph between speed and kinetic energy of a body?
1 A straight line
2 A hyperbola
3 A parabola
4 Exponential
Explanation:
C Let $\mathrm{X}$-axis denotes velocity $\mathrm{v}$ and $\mathrm{Y}$-axis denotes kinetic energy K.E. Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{KE} \propto \mathrm{v}^{2}$ $\mathrm{K} . \mathrm{E}=\mathrm{cv}^{2}$ Where $\mathrm{c}=\frac{1}{2} \mathrm{~m}$ is constant Convert equation (i) in $\mathrm{X}-\mathrm{Y}$ form, $\mathrm{Y}=\mathrm{c} \mathrm{X}^{2}$ So, the graph between speed and kinetic energy of body is parabolic in shape.
AP EAMCET-19.08.2021
Work, Energy and Power
148951
A particle slides down along an inclined plane AC. If the plane is frictionless, kinetic energy of particle at ' $C$ ' is
148948
A stationary object breaks into two pieces $A$ and $B$ of masses $6 \mathrm{~kg}$ and $8 \mathrm{~kg}$ respectively. If the velocity of $B$ is $6 \mathrm{~ms}^{-1}$, then find the kinetic energy of $A$.
1 $48 \mathrm{~J}$
2 $192 \mathrm{~J}$
3 $24 \mathrm{~J}$
4 $288 \mathrm{~J}$
Explanation:
B Given, $\mathrm{m}_{\mathrm{A}}=6 \mathrm{~kg}$ $\mathrm{~m}_{\mathrm{B}}=8 \mathrm{~kg}$ $\mathrm{v}_{\mathrm{B}}=6 \mathrm{~m} / \mathrm{sec}$ By the law of conservation of momentum, $\mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}$ $6 \times \mathrm{v}_{\mathrm{A}}=8 \times 6$ $\mathrm{v}_{\mathrm{A}}=8 \mathrm{~m} / \mathrm{sec}$ Kinetic energy of body A $(\mathrm{K} . \mathrm{E})_{\mathrm{A}} =\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}^{2}$ $=\frac{1}{2} \times 6 \times(8)^{2}$ $=\frac{1}{2} \times 6 \times 64$ $\mathrm{KE}_{\mathrm{A}}= 192 \mathrm{~J}$ The Kinetic Energy of body A is $192 \mathrm{~J}$.
AP EAMCET-06.09.2021
Work, Energy and Power
148949
Two bodies having kinetic energy in the ratio 4 : 1 are moving with same linear velocity. The ratio of their masses is
1 $1: 2$
2 $1: 1$
3 $4: 1$
4 $1: 4$
Explanation:
C Given, $\frac{\mathrm{K} . \mathrm{E}_{1}}{\mathrm{~K} . \mathrm{E}_{2}}=\frac{4}{1}$ We know, $\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{~K} \cdot \mathrm{E} \propto \mathrm{mv}^{2}$ $\frac{\mathrm{K} \cdot \mathrm{E}_{1}}{\mathrm{~K} \cdot \mathrm{E}_{2}}=\frac{\mathrm{m}_{1} \mathrm{v}_{1}^{2}}{\mathrm{~m}_{2} \mathrm{v}_{2}^{2}}$ Given, same linear velocity - $\mathrm{v}_{1}=\mathrm{v}_{2}$ $\frac{4}{1}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}$ The ratio $\mathrm{m}_{1}: \mathrm{m}_{2}=4: 1$
AP EAMCET-20.08.2021
Work, Energy and Power
148950
What is the shape of the graph between speed and kinetic energy of a body?
1 A straight line
2 A hyperbola
3 A parabola
4 Exponential
Explanation:
C Let $\mathrm{X}$-axis denotes velocity $\mathrm{v}$ and $\mathrm{Y}$-axis denotes kinetic energy K.E. Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{KE} \propto \mathrm{v}^{2}$ $\mathrm{K} . \mathrm{E}=\mathrm{cv}^{2}$ Where $\mathrm{c}=\frac{1}{2} \mathrm{~m}$ is constant Convert equation (i) in $\mathrm{X}-\mathrm{Y}$ form, $\mathrm{Y}=\mathrm{c} \mathrm{X}^{2}$ So, the graph between speed and kinetic energy of body is parabolic in shape.
AP EAMCET-19.08.2021
Work, Energy and Power
148951
A particle slides down along an inclined plane AC. If the plane is frictionless, kinetic energy of particle at ' $C$ ' is
148948
A stationary object breaks into two pieces $A$ and $B$ of masses $6 \mathrm{~kg}$ and $8 \mathrm{~kg}$ respectively. If the velocity of $B$ is $6 \mathrm{~ms}^{-1}$, then find the kinetic energy of $A$.
1 $48 \mathrm{~J}$
2 $192 \mathrm{~J}$
3 $24 \mathrm{~J}$
4 $288 \mathrm{~J}$
Explanation:
B Given, $\mathrm{m}_{\mathrm{A}}=6 \mathrm{~kg}$ $\mathrm{~m}_{\mathrm{B}}=8 \mathrm{~kg}$ $\mathrm{v}_{\mathrm{B}}=6 \mathrm{~m} / \mathrm{sec}$ By the law of conservation of momentum, $\mathrm{m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}} \mathrm{v}_{\mathrm{B}}$ $6 \times \mathrm{v}_{\mathrm{A}}=8 \times 6$ $\mathrm{v}_{\mathrm{A}}=8 \mathrm{~m} / \mathrm{sec}$ Kinetic energy of body A $(\mathrm{K} . \mathrm{E})_{\mathrm{A}} =\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{v}_{\mathrm{A}}^{2}$ $=\frac{1}{2} \times 6 \times(8)^{2}$ $=\frac{1}{2} \times 6 \times 64$ $\mathrm{KE}_{\mathrm{A}}= 192 \mathrm{~J}$ The Kinetic Energy of body A is $192 \mathrm{~J}$.
AP EAMCET-06.09.2021
Work, Energy and Power
148949
Two bodies having kinetic energy in the ratio 4 : 1 are moving with same linear velocity. The ratio of their masses is
1 $1: 2$
2 $1: 1$
3 $4: 1$
4 $1: 4$
Explanation:
C Given, $\frac{\mathrm{K} . \mathrm{E}_{1}}{\mathrm{~K} . \mathrm{E}_{2}}=\frac{4}{1}$ We know, $\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{~K} \cdot \mathrm{E} \propto \mathrm{mv}^{2}$ $\frac{\mathrm{K} \cdot \mathrm{E}_{1}}{\mathrm{~K} \cdot \mathrm{E}_{2}}=\frac{\mathrm{m}_{1} \mathrm{v}_{1}^{2}}{\mathrm{~m}_{2} \mathrm{v}_{2}^{2}}$ Given, same linear velocity - $\mathrm{v}_{1}=\mathrm{v}_{2}$ $\frac{4}{1}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}$ The ratio $\mathrm{m}_{1}: \mathrm{m}_{2}=4: 1$
AP EAMCET-20.08.2021
Work, Energy and Power
148950
What is the shape of the graph between speed and kinetic energy of a body?
1 A straight line
2 A hyperbola
3 A parabola
4 Exponential
Explanation:
C Let $\mathrm{X}$-axis denotes velocity $\mathrm{v}$ and $\mathrm{Y}$-axis denotes kinetic energy K.E. Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{KE} \propto \mathrm{v}^{2}$ $\mathrm{K} . \mathrm{E}=\mathrm{cv}^{2}$ Where $\mathrm{c}=\frac{1}{2} \mathrm{~m}$ is constant Convert equation (i) in $\mathrm{X}-\mathrm{Y}$ form, $\mathrm{Y}=\mathrm{c} \mathrm{X}^{2}$ So, the graph between speed and kinetic energy of body is parabolic in shape.
AP EAMCET-19.08.2021
Work, Energy and Power
148951
A particle slides down along an inclined plane AC. If the plane is frictionless, kinetic energy of particle at ' $C$ ' is
