371688
A mass \(m\) falls freely from rest. The linear momentum, after it has fallen through a height \(h\), is : ( \(g\) = acceleration due to gravity)
1 \(\sqrt{\mathrm{mgh}}\)
2 \(\mathrm{m} \sqrt{2 g h}\)
3 \(\mathrm{m} \sqrt{\mathrm{gh}}\)
4 zero
Explanation:
B We know that, Newton's third's law of motion, \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) When initial velocity, \(\mathrm{u}=0\) Then, \(\mathrm{v}^{2} =2 \mathrm{gh}\) \(\mathrm{v} =\sqrt{2 \mathrm{gh}}\) The linear momentum, \(\mathrm{P}=\mathrm{mv}\) \(\mathrm{P}=\mathrm{m} \times \sqrt{2 \mathrm{gh}} \quad\{\because \mathrm{v}=\sqrt{2 \mathrm{gh}}\}\) \(\mathrm{P}=\mathrm{m} \sqrt{2 \mathrm{gh}}\)
AP EAMCET(Medical)-1998
LAWS OF MOTION (ADDITIONAL)
371689
Two masses of \(\mathrm{m}\) and \(4 \mathrm{~m}\) are moving with equal kinetic energy. The ratio of their linear momentum is
371690
The one which does not represent a force in any context is
1 friction
2 impulse
3 tension
4 weight
5 viscous drag
Explanation:
B Impulse is defined as overall effect of a force acting over time and it expressed in Newton-seconds. Impulse \(=\) changes in Momentum \(\therefore\) Impulse is does not represent force.
371688
A mass \(m\) falls freely from rest. The linear momentum, after it has fallen through a height \(h\), is : ( \(g\) = acceleration due to gravity)
1 \(\sqrt{\mathrm{mgh}}\)
2 \(\mathrm{m} \sqrt{2 g h}\)
3 \(\mathrm{m} \sqrt{\mathrm{gh}}\)
4 zero
Explanation:
B We know that, Newton's third's law of motion, \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) When initial velocity, \(\mathrm{u}=0\) Then, \(\mathrm{v}^{2} =2 \mathrm{gh}\) \(\mathrm{v} =\sqrt{2 \mathrm{gh}}\) The linear momentum, \(\mathrm{P}=\mathrm{mv}\) \(\mathrm{P}=\mathrm{m} \times \sqrt{2 \mathrm{gh}} \quad\{\because \mathrm{v}=\sqrt{2 \mathrm{gh}}\}\) \(\mathrm{P}=\mathrm{m} \sqrt{2 \mathrm{gh}}\)
AP EAMCET(Medical)-1998
LAWS OF MOTION (ADDITIONAL)
371689
Two masses of \(\mathrm{m}\) and \(4 \mathrm{~m}\) are moving with equal kinetic energy. The ratio of their linear momentum is
371690
The one which does not represent a force in any context is
1 friction
2 impulse
3 tension
4 weight
5 viscous drag
Explanation:
B Impulse is defined as overall effect of a force acting over time and it expressed in Newton-seconds. Impulse \(=\) changes in Momentum \(\therefore\) Impulse is does not represent force.
371688
A mass \(m\) falls freely from rest. The linear momentum, after it has fallen through a height \(h\), is : ( \(g\) = acceleration due to gravity)
1 \(\sqrt{\mathrm{mgh}}\)
2 \(\mathrm{m} \sqrt{2 g h}\)
3 \(\mathrm{m} \sqrt{\mathrm{gh}}\)
4 zero
Explanation:
B We know that, Newton's third's law of motion, \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) When initial velocity, \(\mathrm{u}=0\) Then, \(\mathrm{v}^{2} =2 \mathrm{gh}\) \(\mathrm{v} =\sqrt{2 \mathrm{gh}}\) The linear momentum, \(\mathrm{P}=\mathrm{mv}\) \(\mathrm{P}=\mathrm{m} \times \sqrt{2 \mathrm{gh}} \quad\{\because \mathrm{v}=\sqrt{2 \mathrm{gh}}\}\) \(\mathrm{P}=\mathrm{m} \sqrt{2 \mathrm{gh}}\)
AP EAMCET(Medical)-1998
LAWS OF MOTION (ADDITIONAL)
371689
Two masses of \(\mathrm{m}\) and \(4 \mathrm{~m}\) are moving with equal kinetic energy. The ratio of their linear momentum is
371690
The one which does not represent a force in any context is
1 friction
2 impulse
3 tension
4 weight
5 viscous drag
Explanation:
B Impulse is defined as overall effect of a force acting over time and it expressed in Newton-seconds. Impulse \(=\) changes in Momentum \(\therefore\) Impulse is does not represent force.
371688
A mass \(m\) falls freely from rest. The linear momentum, after it has fallen through a height \(h\), is : ( \(g\) = acceleration due to gravity)
1 \(\sqrt{\mathrm{mgh}}\)
2 \(\mathrm{m} \sqrt{2 g h}\)
3 \(\mathrm{m} \sqrt{\mathrm{gh}}\)
4 zero
Explanation:
B We know that, Newton's third's law of motion, \(\mathrm{v}^{2}=\mathrm{u}^{2}+2 \mathrm{gh}\) When initial velocity, \(\mathrm{u}=0\) Then, \(\mathrm{v}^{2} =2 \mathrm{gh}\) \(\mathrm{v} =\sqrt{2 \mathrm{gh}}\) The linear momentum, \(\mathrm{P}=\mathrm{mv}\) \(\mathrm{P}=\mathrm{m} \times \sqrt{2 \mathrm{gh}} \quad\{\because \mathrm{v}=\sqrt{2 \mathrm{gh}}\}\) \(\mathrm{P}=\mathrm{m} \sqrt{2 \mathrm{gh}}\)
AP EAMCET(Medical)-1998
LAWS OF MOTION (ADDITIONAL)
371689
Two masses of \(\mathrm{m}\) and \(4 \mathrm{~m}\) are moving with equal kinetic energy. The ratio of their linear momentum is
371690
The one which does not represent a force in any context is
1 friction
2 impulse
3 tension
4 weight
5 viscous drag
Explanation:
B Impulse is defined as overall effect of a force acting over time and it expressed in Newton-seconds. Impulse \(=\) changes in Momentum \(\therefore\) Impulse is does not represent force.
