355226
Two bodies moving towards each other collide and move away in opposite directions. There is some rise in temperature of bodies because a part of the kinetic energy is converted into
1 Electrical energy
2 Heat energy
3 Mechanical energy
4 Nuclear energy
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355227
An object is dropped from a height \(h\) from the ground. Every time it hits the ground it looses \(50 \%\) of its kinetic energy. The total distance covered as \(t \rightarrow \infty\) is
1 2 \(h\)
2 \(\infty\)
3 \(\frac{5}{3}\;h\)
4 3 \(h\)
Explanation:
Final \(K.E\) is \(50 \%\) of initial \(K.E\) \(\begin{aligned}& \dfrac{1}{2} m v=\dfrac{1}{2} \dfrac{1}{2} m u^{2} \\& v=\dfrac{u}{\sqrt{2}} \\& v=e u \Rightarrow e=\dfrac{1}{\sqrt{2}} \\& H=h\left(\dfrac{1+e^{2}}{1-e^{2}}\right)=h\left(\dfrac{1+\dfrac{1}{2}}{1-\dfrac{1}{2}}\right)=3 h\end{aligned}\)
JEE - 2017
PHXI06:WORK ENERGY AND POWER
355228
A small ball moves towards right with a velocity \({v}\) starting from \({A}\). It collides with the wall and returns back and continues to and fro motion. If the average speed for the first trip is \({(2 / 3) v}\), the coefficient of restitution of impact is
1 \(3 \times {10^{ - 1}}\)
2 \(6 \times {10^{ - 1}}\)
3 \(5 \times {10^{ - 1}}\)
4 \(10 \times {10^{ - 1}}\)
Explanation:
The ball moves from \({A}\) to \({B}\) with a speed \({v_{1}=v}\) and returns from \({B}\) to \({A}\) with speed \({v_{2} \Rightarrow v_{2}=e v_{1}}\) \(v_{a v}=\dfrac{A B+B A}{\dfrac{A B}{v_{1}}+\dfrac{B A}{v_{2}}}=\dfrac{2 v}{3}\) \(\left( {AB = BA} \right)\) Solve to get \({e=0.5=5 \times 10^{-1}}\)
PHXI06:WORK ENERGY AND POWER
355229
A moving block having mass \(m\), collides with another stationary block having mass 4 \(m\). The lighter block comes to rest after collision. When the initial velocity of the lighter block is \(v\), then the value of coefficient of restitution (\(e\)) will be
1 0.8
2 0.4
3 0.25
4 0.5
Explanation:
From conservation of linear momentum \(\begin{aligned}& m \cdot v=4 m \cdot v_{1} \\& v_{1}=\dfrac{v}{4}\end{aligned}\) Coefficient of restitution \(e=\dfrac{v_{1}}{v}=\dfrac{1}{4}=0.25\)
355226
Two bodies moving towards each other collide and move away in opposite directions. There is some rise in temperature of bodies because a part of the kinetic energy is converted into
1 Electrical energy
2 Heat energy
3 Mechanical energy
4 Nuclear energy
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355227
An object is dropped from a height \(h\) from the ground. Every time it hits the ground it looses \(50 \%\) of its kinetic energy. The total distance covered as \(t \rightarrow \infty\) is
1 2 \(h\)
2 \(\infty\)
3 \(\frac{5}{3}\;h\)
4 3 \(h\)
Explanation:
Final \(K.E\) is \(50 \%\) of initial \(K.E\) \(\begin{aligned}& \dfrac{1}{2} m v=\dfrac{1}{2} \dfrac{1}{2} m u^{2} \\& v=\dfrac{u}{\sqrt{2}} \\& v=e u \Rightarrow e=\dfrac{1}{\sqrt{2}} \\& H=h\left(\dfrac{1+e^{2}}{1-e^{2}}\right)=h\left(\dfrac{1+\dfrac{1}{2}}{1-\dfrac{1}{2}}\right)=3 h\end{aligned}\)
JEE - 2017
PHXI06:WORK ENERGY AND POWER
355228
A small ball moves towards right with a velocity \({v}\) starting from \({A}\). It collides with the wall and returns back and continues to and fro motion. If the average speed for the first trip is \({(2 / 3) v}\), the coefficient of restitution of impact is
1 \(3 \times {10^{ - 1}}\)
2 \(6 \times {10^{ - 1}}\)
3 \(5 \times {10^{ - 1}}\)
4 \(10 \times {10^{ - 1}}\)
Explanation:
The ball moves from \({A}\) to \({B}\) with a speed \({v_{1}=v}\) and returns from \({B}\) to \({A}\) with speed \({v_{2} \Rightarrow v_{2}=e v_{1}}\) \(v_{a v}=\dfrac{A B+B A}{\dfrac{A B}{v_{1}}+\dfrac{B A}{v_{2}}}=\dfrac{2 v}{3}\) \(\left( {AB = BA} \right)\) Solve to get \({e=0.5=5 \times 10^{-1}}\)
PHXI06:WORK ENERGY AND POWER
355229
A moving block having mass \(m\), collides with another stationary block having mass 4 \(m\). The lighter block comes to rest after collision. When the initial velocity of the lighter block is \(v\), then the value of coefficient of restitution (\(e\)) will be
1 0.8
2 0.4
3 0.25
4 0.5
Explanation:
From conservation of linear momentum \(\begin{aligned}& m \cdot v=4 m \cdot v_{1} \\& v_{1}=\dfrac{v}{4}\end{aligned}\) Coefficient of restitution \(e=\dfrac{v_{1}}{v}=\dfrac{1}{4}=0.25\)
355226
Two bodies moving towards each other collide and move away in opposite directions. There is some rise in temperature of bodies because a part of the kinetic energy is converted into
1 Electrical energy
2 Heat energy
3 Mechanical energy
4 Nuclear energy
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355227
An object is dropped from a height \(h\) from the ground. Every time it hits the ground it looses \(50 \%\) of its kinetic energy. The total distance covered as \(t \rightarrow \infty\) is
1 2 \(h\)
2 \(\infty\)
3 \(\frac{5}{3}\;h\)
4 3 \(h\)
Explanation:
Final \(K.E\) is \(50 \%\) of initial \(K.E\) \(\begin{aligned}& \dfrac{1}{2} m v=\dfrac{1}{2} \dfrac{1}{2} m u^{2} \\& v=\dfrac{u}{\sqrt{2}} \\& v=e u \Rightarrow e=\dfrac{1}{\sqrt{2}} \\& H=h\left(\dfrac{1+e^{2}}{1-e^{2}}\right)=h\left(\dfrac{1+\dfrac{1}{2}}{1-\dfrac{1}{2}}\right)=3 h\end{aligned}\)
JEE - 2017
PHXI06:WORK ENERGY AND POWER
355228
A small ball moves towards right with a velocity \({v}\) starting from \({A}\). It collides with the wall and returns back and continues to and fro motion. If the average speed for the first trip is \({(2 / 3) v}\), the coefficient of restitution of impact is
1 \(3 \times {10^{ - 1}}\)
2 \(6 \times {10^{ - 1}}\)
3 \(5 \times {10^{ - 1}}\)
4 \(10 \times {10^{ - 1}}\)
Explanation:
The ball moves from \({A}\) to \({B}\) with a speed \({v_{1}=v}\) and returns from \({B}\) to \({A}\) with speed \({v_{2} \Rightarrow v_{2}=e v_{1}}\) \(v_{a v}=\dfrac{A B+B A}{\dfrac{A B}{v_{1}}+\dfrac{B A}{v_{2}}}=\dfrac{2 v}{3}\) \(\left( {AB = BA} \right)\) Solve to get \({e=0.5=5 \times 10^{-1}}\)
PHXI06:WORK ENERGY AND POWER
355229
A moving block having mass \(m\), collides with another stationary block having mass 4 \(m\). The lighter block comes to rest after collision. When the initial velocity of the lighter block is \(v\), then the value of coefficient of restitution (\(e\)) will be
1 0.8
2 0.4
3 0.25
4 0.5
Explanation:
From conservation of linear momentum \(\begin{aligned}& m \cdot v=4 m \cdot v_{1} \\& v_{1}=\dfrac{v}{4}\end{aligned}\) Coefficient of restitution \(e=\dfrac{v_{1}}{v}=\dfrac{1}{4}=0.25\)
355226
Two bodies moving towards each other collide and move away in opposite directions. There is some rise in temperature of bodies because a part of the kinetic energy is converted into
1 Electrical energy
2 Heat energy
3 Mechanical energy
4 Nuclear energy
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355227
An object is dropped from a height \(h\) from the ground. Every time it hits the ground it looses \(50 \%\) of its kinetic energy. The total distance covered as \(t \rightarrow \infty\) is
1 2 \(h\)
2 \(\infty\)
3 \(\frac{5}{3}\;h\)
4 3 \(h\)
Explanation:
Final \(K.E\) is \(50 \%\) of initial \(K.E\) \(\begin{aligned}& \dfrac{1}{2} m v=\dfrac{1}{2} \dfrac{1}{2} m u^{2} \\& v=\dfrac{u}{\sqrt{2}} \\& v=e u \Rightarrow e=\dfrac{1}{\sqrt{2}} \\& H=h\left(\dfrac{1+e^{2}}{1-e^{2}}\right)=h\left(\dfrac{1+\dfrac{1}{2}}{1-\dfrac{1}{2}}\right)=3 h\end{aligned}\)
JEE - 2017
PHXI06:WORK ENERGY AND POWER
355228
A small ball moves towards right with a velocity \({v}\) starting from \({A}\). It collides with the wall and returns back and continues to and fro motion. If the average speed for the first trip is \({(2 / 3) v}\), the coefficient of restitution of impact is
1 \(3 \times {10^{ - 1}}\)
2 \(6 \times {10^{ - 1}}\)
3 \(5 \times {10^{ - 1}}\)
4 \(10 \times {10^{ - 1}}\)
Explanation:
The ball moves from \({A}\) to \({B}\) with a speed \({v_{1}=v}\) and returns from \({B}\) to \({A}\) with speed \({v_{2} \Rightarrow v_{2}=e v_{1}}\) \(v_{a v}=\dfrac{A B+B A}{\dfrac{A B}{v_{1}}+\dfrac{B A}{v_{2}}}=\dfrac{2 v}{3}\) \(\left( {AB = BA} \right)\) Solve to get \({e=0.5=5 \times 10^{-1}}\)
PHXI06:WORK ENERGY AND POWER
355229
A moving block having mass \(m\), collides with another stationary block having mass 4 \(m\). The lighter block comes to rest after collision. When the initial velocity of the lighter block is \(v\), then the value of coefficient of restitution (\(e\)) will be
1 0.8
2 0.4
3 0.25
4 0.5
Explanation:
From conservation of linear momentum \(\begin{aligned}& m \cdot v=4 m \cdot v_{1} \\& v_{1}=\dfrac{v}{4}\end{aligned}\) Coefficient of restitution \(e=\dfrac{v_{1}}{v}=\dfrac{1}{4}=0.25\)
