355222
A body falls on a surface of coefficient of restitution 0.6 from a height of 1 \(m\). Then the body rebounds to a height of
1 0.4 \(m\)
2 0.36 \(m\)
3 1 \(m\)
4 0.6 \(m\)
Explanation:
The rebouncing height after \(n^{\text {th }}\) collision is \(h_{n}=h e^{2 n}=1 \times e^{2}=1 \times(0.6)^{2}=0.36 m\)
PHXI06:WORK ENERGY AND POWER
355223
A body is allowed to fall on the ground from a height \(h_{1}\). If it rebounds to a height \(h_{2}\) then the coefficient of restitution is:
1 \(\sqrt{\dfrac{h_{2}}{h_{1}}}\)
2 \(\dfrac{h_{2}}{h_{1}}\)
3 \(\sqrt{\dfrac{h_{1}}{h_{2}}}\)
4 \(\sqrt{\dfrac{h_{1}}{h_{2}}}\)
Explanation:
\(u_{1}=-\sqrt{2 g h_{1}}, v_{1}=\sqrt{2 g h_{2}}\) We know that \(v_{1}=e u\) \(\Rightarrow e=\dfrac{v_{1}}{u_{1}}=\sqrt{\dfrac{h_{2}}{h_{1}}}\)
PHXI06:WORK ENERGY AND POWER
355224
A bullet of mass \(20\;g\) and moving with \(600\;m{s^{ - 1}}\) collides with a block of mass \(4\;kg\) hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height \(0.2 \mathrm{~m}\) after collision?
1 \(200\;m{s^{ - 1}}\)
2 \(150\;m{s^{ - 1}}\)
3 \(400\;m{s^{ - 1}}\)
4 \(300\;m{s^{ - 1}}\)
Explanation:
According to the law of conservation of linear momentum, \(m_{1} v=m_{1} v_{1}+m_{2} v_{2}\) where, \(v\) is the velocity of bullet before the collision, \(v_{1}\) is velocity of the bullet after the collision and \(v_{2}\) is the velocity of block. Given, \({m_1} = 20g = 0.02\;kg,\) \(v = 600\;m{s^{ - 1}},\,\,{m_2} = 4\;kg\) and \(h = 0.2\;m\) \(\therefore \quad 0.02 \times 600 = 0.02{v_1} + 4{v_2}\) \(\left( {\because {v_2} = \sqrt {2gh} = \sqrt {2 \times 10 \times 0.2} = 2\;m{s^{ - 1}}} \right)\) \( \Rightarrow \quad 0.02\,\,{v_1} = 12 - 8\) \( \Rightarrow \quad {v_1} = \frac{4}{{0.02}} = 200\;m{s^{ - 1}}\)
PHXI06:WORK ENERGY AND POWER
355225
A sphere of mass \(m\) moving with a constant velocity \(u\) hits another stationary sphere of same mass. if \(e\) is the coefficient of restitution, the ratio of velocities of the two spheres \(\dfrac{v_{1}}{v_{2}}\) after collision will be
1 \(\dfrac{1-e}{1+e}\)
2 \(\dfrac{1+e}{1-e}\)
3 \(\dfrac{e+1}{e-1}\)
4 \(\dfrac{e-1}{e+1}\)
Explanation:
From conservation of linear momentum \(m u=m v_{1}+m v_{2}\) or \(u=v_{1}+v_{2}\) From definition of \(e\): \({v_1} - {v_2} = eu\) Solving these two equations, we get \({v_1} = \left( {\frac{{1 + e}}{2}} \right)u\) and \({v_2} = \left( {\frac{{1 - e}}{2}} \right)u\) \(\therefore \frac{{{v_1}}}{{{v_2}}} = \left( {\frac{{1 + e}}{{1 - e}}} \right)\)
355222
A body falls on a surface of coefficient of restitution 0.6 from a height of 1 \(m\). Then the body rebounds to a height of
1 0.4 \(m\)
2 0.36 \(m\)
3 1 \(m\)
4 0.6 \(m\)
Explanation:
The rebouncing height after \(n^{\text {th }}\) collision is \(h_{n}=h e^{2 n}=1 \times e^{2}=1 \times(0.6)^{2}=0.36 m\)
PHXI06:WORK ENERGY AND POWER
355223
A body is allowed to fall on the ground from a height \(h_{1}\). If it rebounds to a height \(h_{2}\) then the coefficient of restitution is:
1 \(\sqrt{\dfrac{h_{2}}{h_{1}}}\)
2 \(\dfrac{h_{2}}{h_{1}}\)
3 \(\sqrt{\dfrac{h_{1}}{h_{2}}}\)
4 \(\sqrt{\dfrac{h_{1}}{h_{2}}}\)
Explanation:
\(u_{1}=-\sqrt{2 g h_{1}}, v_{1}=\sqrt{2 g h_{2}}\) We know that \(v_{1}=e u\) \(\Rightarrow e=\dfrac{v_{1}}{u_{1}}=\sqrt{\dfrac{h_{2}}{h_{1}}}\)
PHXI06:WORK ENERGY AND POWER
355224
A bullet of mass \(20\;g\) and moving with \(600\;m{s^{ - 1}}\) collides with a block of mass \(4\;kg\) hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height \(0.2 \mathrm{~m}\) after collision?
1 \(200\;m{s^{ - 1}}\)
2 \(150\;m{s^{ - 1}}\)
3 \(400\;m{s^{ - 1}}\)
4 \(300\;m{s^{ - 1}}\)
Explanation:
According to the law of conservation of linear momentum, \(m_{1} v=m_{1} v_{1}+m_{2} v_{2}\) where, \(v\) is the velocity of bullet before the collision, \(v_{1}\) is velocity of the bullet after the collision and \(v_{2}\) is the velocity of block. Given, \({m_1} = 20g = 0.02\;kg,\) \(v = 600\;m{s^{ - 1}},\,\,{m_2} = 4\;kg\) and \(h = 0.2\;m\) \(\therefore \quad 0.02 \times 600 = 0.02{v_1} + 4{v_2}\) \(\left( {\because {v_2} = \sqrt {2gh} = \sqrt {2 \times 10 \times 0.2} = 2\;m{s^{ - 1}}} \right)\) \( \Rightarrow \quad 0.02\,\,{v_1} = 12 - 8\) \( \Rightarrow \quad {v_1} = \frac{4}{{0.02}} = 200\;m{s^{ - 1}}\)
PHXI06:WORK ENERGY AND POWER
355225
A sphere of mass \(m\) moving with a constant velocity \(u\) hits another stationary sphere of same mass. if \(e\) is the coefficient of restitution, the ratio of velocities of the two spheres \(\dfrac{v_{1}}{v_{2}}\) after collision will be
1 \(\dfrac{1-e}{1+e}\)
2 \(\dfrac{1+e}{1-e}\)
3 \(\dfrac{e+1}{e-1}\)
4 \(\dfrac{e-1}{e+1}\)
Explanation:
From conservation of linear momentum \(m u=m v_{1}+m v_{2}\) or \(u=v_{1}+v_{2}\) From definition of \(e\): \({v_1} - {v_2} = eu\) Solving these two equations, we get \({v_1} = \left( {\frac{{1 + e}}{2}} \right)u\) and \({v_2} = \left( {\frac{{1 - e}}{2}} \right)u\) \(\therefore \frac{{{v_1}}}{{{v_2}}} = \left( {\frac{{1 + e}}{{1 - e}}} \right)\)
355222
A body falls on a surface of coefficient of restitution 0.6 from a height of 1 \(m\). Then the body rebounds to a height of
1 0.4 \(m\)
2 0.36 \(m\)
3 1 \(m\)
4 0.6 \(m\)
Explanation:
The rebouncing height after \(n^{\text {th }}\) collision is \(h_{n}=h e^{2 n}=1 \times e^{2}=1 \times(0.6)^{2}=0.36 m\)
PHXI06:WORK ENERGY AND POWER
355223
A body is allowed to fall on the ground from a height \(h_{1}\). If it rebounds to a height \(h_{2}\) then the coefficient of restitution is:
1 \(\sqrt{\dfrac{h_{2}}{h_{1}}}\)
2 \(\dfrac{h_{2}}{h_{1}}\)
3 \(\sqrt{\dfrac{h_{1}}{h_{2}}}\)
4 \(\sqrt{\dfrac{h_{1}}{h_{2}}}\)
Explanation:
\(u_{1}=-\sqrt{2 g h_{1}}, v_{1}=\sqrt{2 g h_{2}}\) We know that \(v_{1}=e u\) \(\Rightarrow e=\dfrac{v_{1}}{u_{1}}=\sqrt{\dfrac{h_{2}}{h_{1}}}\)
PHXI06:WORK ENERGY AND POWER
355224
A bullet of mass \(20\;g\) and moving with \(600\;m{s^{ - 1}}\) collides with a block of mass \(4\;kg\) hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height \(0.2 \mathrm{~m}\) after collision?
1 \(200\;m{s^{ - 1}}\)
2 \(150\;m{s^{ - 1}}\)
3 \(400\;m{s^{ - 1}}\)
4 \(300\;m{s^{ - 1}}\)
Explanation:
According to the law of conservation of linear momentum, \(m_{1} v=m_{1} v_{1}+m_{2} v_{2}\) where, \(v\) is the velocity of bullet before the collision, \(v_{1}\) is velocity of the bullet after the collision and \(v_{2}\) is the velocity of block. Given, \({m_1} = 20g = 0.02\;kg,\) \(v = 600\;m{s^{ - 1}},\,\,{m_2} = 4\;kg\) and \(h = 0.2\;m\) \(\therefore \quad 0.02 \times 600 = 0.02{v_1} + 4{v_2}\) \(\left( {\because {v_2} = \sqrt {2gh} = \sqrt {2 \times 10 \times 0.2} = 2\;m{s^{ - 1}}} \right)\) \( \Rightarrow \quad 0.02\,\,{v_1} = 12 - 8\) \( \Rightarrow \quad {v_1} = \frac{4}{{0.02}} = 200\;m{s^{ - 1}}\)
PHXI06:WORK ENERGY AND POWER
355225
A sphere of mass \(m\) moving with a constant velocity \(u\) hits another stationary sphere of same mass. if \(e\) is the coefficient of restitution, the ratio of velocities of the two spheres \(\dfrac{v_{1}}{v_{2}}\) after collision will be
1 \(\dfrac{1-e}{1+e}\)
2 \(\dfrac{1+e}{1-e}\)
3 \(\dfrac{e+1}{e-1}\)
4 \(\dfrac{e-1}{e+1}\)
Explanation:
From conservation of linear momentum \(m u=m v_{1}+m v_{2}\) or \(u=v_{1}+v_{2}\) From definition of \(e\): \({v_1} - {v_2} = eu\) Solving these two equations, we get \({v_1} = \left( {\frac{{1 + e}}{2}} \right)u\) and \({v_2} = \left( {\frac{{1 - e}}{2}} \right)u\) \(\therefore \frac{{{v_1}}}{{{v_2}}} = \left( {\frac{{1 + e}}{{1 - e}}} \right)\)
355222
A body falls on a surface of coefficient of restitution 0.6 from a height of 1 \(m\). Then the body rebounds to a height of
1 0.4 \(m\)
2 0.36 \(m\)
3 1 \(m\)
4 0.6 \(m\)
Explanation:
The rebouncing height after \(n^{\text {th }}\) collision is \(h_{n}=h e^{2 n}=1 \times e^{2}=1 \times(0.6)^{2}=0.36 m\)
PHXI06:WORK ENERGY AND POWER
355223
A body is allowed to fall on the ground from a height \(h_{1}\). If it rebounds to a height \(h_{2}\) then the coefficient of restitution is:
1 \(\sqrt{\dfrac{h_{2}}{h_{1}}}\)
2 \(\dfrac{h_{2}}{h_{1}}\)
3 \(\sqrt{\dfrac{h_{1}}{h_{2}}}\)
4 \(\sqrt{\dfrac{h_{1}}{h_{2}}}\)
Explanation:
\(u_{1}=-\sqrt{2 g h_{1}}, v_{1}=\sqrt{2 g h_{2}}\) We know that \(v_{1}=e u\) \(\Rightarrow e=\dfrac{v_{1}}{u_{1}}=\sqrt{\dfrac{h_{2}}{h_{1}}}\)
PHXI06:WORK ENERGY AND POWER
355224
A bullet of mass \(20\;g\) and moving with \(600\;m{s^{ - 1}}\) collides with a block of mass \(4\;kg\) hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height \(0.2 \mathrm{~m}\) after collision?
1 \(200\;m{s^{ - 1}}\)
2 \(150\;m{s^{ - 1}}\)
3 \(400\;m{s^{ - 1}}\)
4 \(300\;m{s^{ - 1}}\)
Explanation:
According to the law of conservation of linear momentum, \(m_{1} v=m_{1} v_{1}+m_{2} v_{2}\) where, \(v\) is the velocity of bullet before the collision, \(v_{1}\) is velocity of the bullet after the collision and \(v_{2}\) is the velocity of block. Given, \({m_1} = 20g = 0.02\;kg,\) \(v = 600\;m{s^{ - 1}},\,\,{m_2} = 4\;kg\) and \(h = 0.2\;m\) \(\therefore \quad 0.02 \times 600 = 0.02{v_1} + 4{v_2}\) \(\left( {\because {v_2} = \sqrt {2gh} = \sqrt {2 \times 10 \times 0.2} = 2\;m{s^{ - 1}}} \right)\) \( \Rightarrow \quad 0.02\,\,{v_1} = 12 - 8\) \( \Rightarrow \quad {v_1} = \frac{4}{{0.02}} = 200\;m{s^{ - 1}}\)
PHXI06:WORK ENERGY AND POWER
355225
A sphere of mass \(m\) moving with a constant velocity \(u\) hits another stationary sphere of same mass. if \(e\) is the coefficient of restitution, the ratio of velocities of the two spheres \(\dfrac{v_{1}}{v_{2}}\) after collision will be
1 \(\dfrac{1-e}{1+e}\)
2 \(\dfrac{1+e}{1-e}\)
3 \(\dfrac{e+1}{e-1}\)
4 \(\dfrac{e-1}{e+1}\)
Explanation:
From conservation of linear momentum \(m u=m v_{1}+m v_{2}\) or \(u=v_{1}+v_{2}\) From definition of \(e\): \({v_1} - {v_2} = eu\) Solving these two equations, we get \({v_1} = \left( {\frac{{1 + e}}{2}} \right)u\) and \({v_2} = \left( {\frac{{1 - e}}{2}} \right)u\) \(\therefore \frac{{{v_1}}}{{{v_2}}} = \left( {\frac{{1 + e}}{{1 - e}}} \right)\)
