355273
A ball hits a floor and rebounds after an elastic collision. In this case:
1 The kinetic energy of the ball remains the same during the collision.
2 The momentum of the ball just after the collision is same as that just before the collision
3 The total momentum of the ball and the earth is conserved
4 None of these
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355274
An elevator platform is going up at a speed \(20 {~ms}^{-1}\) and during its upward motion a small ball of 50 \(g\) mass falling in downward direction strikes the platform elastically at a speed \(5 {~ms}^{-1}\). Find the speed with which the ball rebounds.
1 \(32\,m/{s^{ - 1}}\)
2 \(55\,\,m/{s^{ - 1}}\)
3 \(45\,m/{s^{ - 1}}\)
4 \(28\,m/{s^{ - 1}}\)
Explanation:
The situation is analysed in the figure. We can consider mass of platform to be very large compared to that of ball, so we have \({\vec v_1} = {\vec u_1}e = \left( {\frac{{{v_2} - {v_1}}}{{{u_2} - {u_1}}}} \right) = \left( {\frac{{{v_2} - {u_1}}}{{{u_1} - \left( { - {u_2}} \right)}}} \right)\) \(\vec{v}_{2}=2 \vec{u}_{1}-\vec{u}_{2}\) Thus rebound velocity of ball is \(\vec{v}_{2}=2 \times 20-(-5)=45 {~ms}^{-1} \text { upward }\)
PHXI06:WORK ENERGY AND POWER
355275
A ball mass 1 \(kg\) stirkes a heavy platform, elastically, moving upwards with a velocity of 5 \(m/s\). The speed of the ball just before the collision is 10 \(m/s\) downwards. The impluse imparted by the platform on the ball is
1 \(10\;N - s\)
2 \(15\;N - s\)
3 \(30\,N - s\)
4 \(20\,N - s\)
Explanation:
Let is the velocity of the ball after collision. Velocity of the platform remains constant. As we know \(e=\dfrac{\left(v_{2}-v_{1}\right)}{\left(u_{1}-u_{2}\right)}\) \(1 = \frac{{( - 5) - \left( {{v_1}} \right)}}{{(10) - ( - 5)}} \Rightarrow {v_1} = - 20\;m/s\) Hence ball will start moving towards upward direction with velocity \(20 \mathrm{~m} / \mathrm{s}\). Let impulse imparted by plate force on the ball is \(J\) in upward direction. \(\,\hat j = {\overrightarrow P _f} - {\overrightarrow P _i} = \left( 1 \right)\left( { - 20} \right)\hat j - \left( 1 \right)10\hat j\) \( \Rightarrow \quad J = 30\,kg\,m/s\)
PHXI06:WORK ENERGY AND POWER
355276
Body of mass \(M\) is much heavier than the other body of mass \(m\). The heavier body with speed \(v\) collides with the lighter body which was at rest initially elastically. The speed of lighter body after collision is
1 \(2 v\)
2 \(3 v\)
3 \(v\)
4 \(\dfrac{v}{2}\)
Explanation:
If \(m_{1}\) and \(m_{2}\) collide elastically we know that \(\begin{aligned}& v_{2}=\dfrac{2 m_{1}}{m_{1}+m_{2}} v_{1}+\dfrac{m_{2}-m_{1}}{m_{1}+m_{2}} v_{2} \\& \therefore v_{2}=0 \\& v_{2}=\dfrac{2 m_{1}}{m_{1}+m_{2}} v_{1} \\& =\dfrac{2 M}{M+m} v_{1} \\& \approx 2 v_{1}(\because M>>>m)\end{aligned}\)
355273
A ball hits a floor and rebounds after an elastic collision. In this case:
1 The kinetic energy of the ball remains the same during the collision.
2 The momentum of the ball just after the collision is same as that just before the collision
3 The total momentum of the ball and the earth is conserved
4 None of these
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355274
An elevator platform is going up at a speed \(20 {~ms}^{-1}\) and during its upward motion a small ball of 50 \(g\) mass falling in downward direction strikes the platform elastically at a speed \(5 {~ms}^{-1}\). Find the speed with which the ball rebounds.
1 \(32\,m/{s^{ - 1}}\)
2 \(55\,\,m/{s^{ - 1}}\)
3 \(45\,m/{s^{ - 1}}\)
4 \(28\,m/{s^{ - 1}}\)
Explanation:
The situation is analysed in the figure. We can consider mass of platform to be very large compared to that of ball, so we have \({\vec v_1} = {\vec u_1}e = \left( {\frac{{{v_2} - {v_1}}}{{{u_2} - {u_1}}}} \right) = \left( {\frac{{{v_2} - {u_1}}}{{{u_1} - \left( { - {u_2}} \right)}}} \right)\) \(\vec{v}_{2}=2 \vec{u}_{1}-\vec{u}_{2}\) Thus rebound velocity of ball is \(\vec{v}_{2}=2 \times 20-(-5)=45 {~ms}^{-1} \text { upward }\)
PHXI06:WORK ENERGY AND POWER
355275
A ball mass 1 \(kg\) stirkes a heavy platform, elastically, moving upwards with a velocity of 5 \(m/s\). The speed of the ball just before the collision is 10 \(m/s\) downwards. The impluse imparted by the platform on the ball is
1 \(10\;N - s\)
2 \(15\;N - s\)
3 \(30\,N - s\)
4 \(20\,N - s\)
Explanation:
Let is the velocity of the ball after collision. Velocity of the platform remains constant. As we know \(e=\dfrac{\left(v_{2}-v_{1}\right)}{\left(u_{1}-u_{2}\right)}\) \(1 = \frac{{( - 5) - \left( {{v_1}} \right)}}{{(10) - ( - 5)}} \Rightarrow {v_1} = - 20\;m/s\) Hence ball will start moving towards upward direction with velocity \(20 \mathrm{~m} / \mathrm{s}\). Let impulse imparted by plate force on the ball is \(J\) in upward direction. \(\,\hat j = {\overrightarrow P _f} - {\overrightarrow P _i} = \left( 1 \right)\left( { - 20} \right)\hat j - \left( 1 \right)10\hat j\) \( \Rightarrow \quad J = 30\,kg\,m/s\)
PHXI06:WORK ENERGY AND POWER
355276
Body of mass \(M\) is much heavier than the other body of mass \(m\). The heavier body with speed \(v\) collides with the lighter body which was at rest initially elastically. The speed of lighter body after collision is
1 \(2 v\)
2 \(3 v\)
3 \(v\)
4 \(\dfrac{v}{2}\)
Explanation:
If \(m_{1}\) and \(m_{2}\) collide elastically we know that \(\begin{aligned}& v_{2}=\dfrac{2 m_{1}}{m_{1}+m_{2}} v_{1}+\dfrac{m_{2}-m_{1}}{m_{1}+m_{2}} v_{2} \\& \therefore v_{2}=0 \\& v_{2}=\dfrac{2 m_{1}}{m_{1}+m_{2}} v_{1} \\& =\dfrac{2 M}{M+m} v_{1} \\& \approx 2 v_{1}(\because M>>>m)\end{aligned}\)
355273
A ball hits a floor and rebounds after an elastic collision. In this case:
1 The kinetic energy of the ball remains the same during the collision.
2 The momentum of the ball just after the collision is same as that just before the collision
3 The total momentum of the ball and the earth is conserved
4 None of these
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355274
An elevator platform is going up at a speed \(20 {~ms}^{-1}\) and during its upward motion a small ball of 50 \(g\) mass falling in downward direction strikes the platform elastically at a speed \(5 {~ms}^{-1}\). Find the speed with which the ball rebounds.
1 \(32\,m/{s^{ - 1}}\)
2 \(55\,\,m/{s^{ - 1}}\)
3 \(45\,m/{s^{ - 1}}\)
4 \(28\,m/{s^{ - 1}}\)
Explanation:
The situation is analysed in the figure. We can consider mass of platform to be very large compared to that of ball, so we have \({\vec v_1} = {\vec u_1}e = \left( {\frac{{{v_2} - {v_1}}}{{{u_2} - {u_1}}}} \right) = \left( {\frac{{{v_2} - {u_1}}}{{{u_1} - \left( { - {u_2}} \right)}}} \right)\) \(\vec{v}_{2}=2 \vec{u}_{1}-\vec{u}_{2}\) Thus rebound velocity of ball is \(\vec{v}_{2}=2 \times 20-(-5)=45 {~ms}^{-1} \text { upward }\)
PHXI06:WORK ENERGY AND POWER
355275
A ball mass 1 \(kg\) stirkes a heavy platform, elastically, moving upwards with a velocity of 5 \(m/s\). The speed of the ball just before the collision is 10 \(m/s\) downwards. The impluse imparted by the platform on the ball is
1 \(10\;N - s\)
2 \(15\;N - s\)
3 \(30\,N - s\)
4 \(20\,N - s\)
Explanation:
Let is the velocity of the ball after collision. Velocity of the platform remains constant. As we know \(e=\dfrac{\left(v_{2}-v_{1}\right)}{\left(u_{1}-u_{2}\right)}\) \(1 = \frac{{( - 5) - \left( {{v_1}} \right)}}{{(10) - ( - 5)}} \Rightarrow {v_1} = - 20\;m/s\) Hence ball will start moving towards upward direction with velocity \(20 \mathrm{~m} / \mathrm{s}\). Let impulse imparted by plate force on the ball is \(J\) in upward direction. \(\,\hat j = {\overrightarrow P _f} - {\overrightarrow P _i} = \left( 1 \right)\left( { - 20} \right)\hat j - \left( 1 \right)10\hat j\) \( \Rightarrow \quad J = 30\,kg\,m/s\)
PHXI06:WORK ENERGY AND POWER
355276
Body of mass \(M\) is much heavier than the other body of mass \(m\). The heavier body with speed \(v\) collides with the lighter body which was at rest initially elastically. The speed of lighter body after collision is
1 \(2 v\)
2 \(3 v\)
3 \(v\)
4 \(\dfrac{v}{2}\)
Explanation:
If \(m_{1}\) and \(m_{2}\) collide elastically we know that \(\begin{aligned}& v_{2}=\dfrac{2 m_{1}}{m_{1}+m_{2}} v_{1}+\dfrac{m_{2}-m_{1}}{m_{1}+m_{2}} v_{2} \\& \therefore v_{2}=0 \\& v_{2}=\dfrac{2 m_{1}}{m_{1}+m_{2}} v_{1} \\& =\dfrac{2 M}{M+m} v_{1} \\& \approx 2 v_{1}(\because M>>>m)\end{aligned}\)
355273
A ball hits a floor and rebounds after an elastic collision. In this case:
1 The kinetic energy of the ball remains the same during the collision.
2 The momentum of the ball just after the collision is same as that just before the collision
3 The total momentum of the ball and the earth is conserved
4 None of these
Explanation:
Conceptual Question
PHXI06:WORK ENERGY AND POWER
355274
An elevator platform is going up at a speed \(20 {~ms}^{-1}\) and during its upward motion a small ball of 50 \(g\) mass falling in downward direction strikes the platform elastically at a speed \(5 {~ms}^{-1}\). Find the speed with which the ball rebounds.
1 \(32\,m/{s^{ - 1}}\)
2 \(55\,\,m/{s^{ - 1}}\)
3 \(45\,m/{s^{ - 1}}\)
4 \(28\,m/{s^{ - 1}}\)
Explanation:
The situation is analysed in the figure. We can consider mass of platform to be very large compared to that of ball, so we have \({\vec v_1} = {\vec u_1}e = \left( {\frac{{{v_2} - {v_1}}}{{{u_2} - {u_1}}}} \right) = \left( {\frac{{{v_2} - {u_1}}}{{{u_1} - \left( { - {u_2}} \right)}}} \right)\) \(\vec{v}_{2}=2 \vec{u}_{1}-\vec{u}_{2}\) Thus rebound velocity of ball is \(\vec{v}_{2}=2 \times 20-(-5)=45 {~ms}^{-1} \text { upward }\)
PHXI06:WORK ENERGY AND POWER
355275
A ball mass 1 \(kg\) stirkes a heavy platform, elastically, moving upwards with a velocity of 5 \(m/s\). The speed of the ball just before the collision is 10 \(m/s\) downwards. The impluse imparted by the platform on the ball is
1 \(10\;N - s\)
2 \(15\;N - s\)
3 \(30\,N - s\)
4 \(20\,N - s\)
Explanation:
Let is the velocity of the ball after collision. Velocity of the platform remains constant. As we know \(e=\dfrac{\left(v_{2}-v_{1}\right)}{\left(u_{1}-u_{2}\right)}\) \(1 = \frac{{( - 5) - \left( {{v_1}} \right)}}{{(10) - ( - 5)}} \Rightarrow {v_1} = - 20\;m/s\) Hence ball will start moving towards upward direction with velocity \(20 \mathrm{~m} / \mathrm{s}\). Let impulse imparted by plate force on the ball is \(J\) in upward direction. \(\,\hat j = {\overrightarrow P _f} - {\overrightarrow P _i} = \left( 1 \right)\left( { - 20} \right)\hat j - \left( 1 \right)10\hat j\) \( \Rightarrow \quad J = 30\,kg\,m/s\)
PHXI06:WORK ENERGY AND POWER
355276
Body of mass \(M\) is much heavier than the other body of mass \(m\). The heavier body with speed \(v\) collides with the lighter body which was at rest initially elastically. The speed of lighter body after collision is
1 \(2 v\)
2 \(3 v\)
3 \(v\)
4 \(\dfrac{v}{2}\)
Explanation:
If \(m_{1}\) and \(m_{2}\) collide elastically we know that \(\begin{aligned}& v_{2}=\dfrac{2 m_{1}}{m_{1}+m_{2}} v_{1}+\dfrac{m_{2}-m_{1}}{m_{1}+m_{2}} v_{2} \\& \therefore v_{2}=0 \\& v_{2}=\dfrac{2 m_{1}}{m_{1}+m_{2}} v_{1} \\& =\dfrac{2 M}{M+m} v_{1} \\& \approx 2 v_{1}(\because M>>>m)\end{aligned}\)
