363493
Assertion : Even though net external force on a body is zero, momentum need not be conserved. Reason : The internal interaction between particles of a body cancels out momentum of each other.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and reason are incorrect.
Explanation:
Since \({\vec F_{ext{\rm{ }}}} = 0 \Rightarrow \frac{{d\vec p}}{{dt}} = 0\) \(\Rightarrow\) momentum gets conserved. Also the internal interactions involve internal forces. Internal forces cancel each other not the linear momenta.
AIIMS - 2019
PHXI05:LAWS OF MOTION
363494
Statement A : A body is momentarily at rest but still some force is acting on it at that time. Statement B : When a force acts on a body, it may not have some acceleration.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
A stationary body \((v = 0)\) may still have some acceleration, e.g. when a body is thrown in upward direction, it comes to rest at highest position, but at that time, it still have acceleration equal to acceleration due to gravity \(g\). Hence, gravitational force is acting at highest position and when a force acts on a body, then its accelerates. Therefore, Statement A is correct but Statement B is incorrect.
PHXI05:LAWS OF MOTION
363495
A gas molecule of mass \(m\) strikes the wall of the container with a speed \(v\) at an angle \(\theta \) with the normal to the wall at the point of collision. The impluse of the gas molecule has a magnitude
1 \(2\,mV\cos \theta \)
2 \(3\,mV\)
3 \({\rm{Zero}}\)
4 \(mV\)
Explanation:
From adjoining figure the component of momentum along \(x\)-axis (parallel to the wall of container) remains unchanged even after the collision. \(\therefore \) Impluse \(=\) change in momentum of gas molecule along \(y\)-axis, \( = 2mv\cos \theta \)
PHXI05:LAWS OF MOTION
363496
A body of mass \(4\,kg\) starts from rest, and a force is acting on the body, as shown in the figure. What is the speed of the body after \({4 s}\) ?
1 \({15 {~m} / {s}}\)
2 \({20 {~m} / {s}}\)
3 \({5 {~m} / {s}}\)
4 \({25 {~m} / {s}}\)
Explanation:
Area under graph, \(\begin{aligned}\Delta p & =\dfrac{1}{2} \times F \times t \\& =\dfrac{1}{2} \times 4 \times 10=20 {~kg} {~m} {~s}^{-1} \\\Delta p & =(m v-m u) \Rightarrow 20=m(v-u) \\20 & =4(v-0) v=5 {~ms}^{-1}\end{aligned}\)
363493
Assertion : Even though net external force on a body is zero, momentum need not be conserved. Reason : The internal interaction between particles of a body cancels out momentum of each other.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and reason are incorrect.
Explanation:
Since \({\vec F_{ext{\rm{ }}}} = 0 \Rightarrow \frac{{d\vec p}}{{dt}} = 0\) \(\Rightarrow\) momentum gets conserved. Also the internal interactions involve internal forces. Internal forces cancel each other not the linear momenta.
AIIMS - 2019
PHXI05:LAWS OF MOTION
363494
Statement A : A body is momentarily at rest but still some force is acting on it at that time. Statement B : When a force acts on a body, it may not have some acceleration.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
A stationary body \((v = 0)\) may still have some acceleration, e.g. when a body is thrown in upward direction, it comes to rest at highest position, but at that time, it still have acceleration equal to acceleration due to gravity \(g\). Hence, gravitational force is acting at highest position and when a force acts on a body, then its accelerates. Therefore, Statement A is correct but Statement B is incorrect.
PHXI05:LAWS OF MOTION
363495
A gas molecule of mass \(m\) strikes the wall of the container with a speed \(v\) at an angle \(\theta \) with the normal to the wall at the point of collision. The impluse of the gas molecule has a magnitude
1 \(2\,mV\cos \theta \)
2 \(3\,mV\)
3 \({\rm{Zero}}\)
4 \(mV\)
Explanation:
From adjoining figure the component of momentum along \(x\)-axis (parallel to the wall of container) remains unchanged even after the collision. \(\therefore \) Impluse \(=\) change in momentum of gas molecule along \(y\)-axis, \( = 2mv\cos \theta \)
PHXI05:LAWS OF MOTION
363496
A body of mass \(4\,kg\) starts from rest, and a force is acting on the body, as shown in the figure. What is the speed of the body after \({4 s}\) ?
1 \({15 {~m} / {s}}\)
2 \({20 {~m} / {s}}\)
3 \({5 {~m} / {s}}\)
4 \({25 {~m} / {s}}\)
Explanation:
Area under graph, \(\begin{aligned}\Delta p & =\dfrac{1}{2} \times F \times t \\& =\dfrac{1}{2} \times 4 \times 10=20 {~kg} {~m} {~s}^{-1} \\\Delta p & =(m v-m u) \Rightarrow 20=m(v-u) \\20 & =4(v-0) v=5 {~ms}^{-1}\end{aligned}\)
363493
Assertion : Even though net external force on a body is zero, momentum need not be conserved. Reason : The internal interaction between particles of a body cancels out momentum of each other.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and reason are incorrect.
Explanation:
Since \({\vec F_{ext{\rm{ }}}} = 0 \Rightarrow \frac{{d\vec p}}{{dt}} = 0\) \(\Rightarrow\) momentum gets conserved. Also the internal interactions involve internal forces. Internal forces cancel each other not the linear momenta.
AIIMS - 2019
PHXI05:LAWS OF MOTION
363494
Statement A : A body is momentarily at rest but still some force is acting on it at that time. Statement B : When a force acts on a body, it may not have some acceleration.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
A stationary body \((v = 0)\) may still have some acceleration, e.g. when a body is thrown in upward direction, it comes to rest at highest position, but at that time, it still have acceleration equal to acceleration due to gravity \(g\). Hence, gravitational force is acting at highest position and when a force acts on a body, then its accelerates. Therefore, Statement A is correct but Statement B is incorrect.
PHXI05:LAWS OF MOTION
363495
A gas molecule of mass \(m\) strikes the wall of the container with a speed \(v\) at an angle \(\theta \) with the normal to the wall at the point of collision. The impluse of the gas molecule has a magnitude
1 \(2\,mV\cos \theta \)
2 \(3\,mV\)
3 \({\rm{Zero}}\)
4 \(mV\)
Explanation:
From adjoining figure the component of momentum along \(x\)-axis (parallel to the wall of container) remains unchanged even after the collision. \(\therefore \) Impluse \(=\) change in momentum of gas molecule along \(y\)-axis, \( = 2mv\cos \theta \)
PHXI05:LAWS OF MOTION
363496
A body of mass \(4\,kg\) starts from rest, and a force is acting on the body, as shown in the figure. What is the speed of the body after \({4 s}\) ?
1 \({15 {~m} / {s}}\)
2 \({20 {~m} / {s}}\)
3 \({5 {~m} / {s}}\)
4 \({25 {~m} / {s}}\)
Explanation:
Area under graph, \(\begin{aligned}\Delta p & =\dfrac{1}{2} \times F \times t \\& =\dfrac{1}{2} \times 4 \times 10=20 {~kg} {~m} {~s}^{-1} \\\Delta p & =(m v-m u) \Rightarrow 20=m(v-u) \\20 & =4(v-0) v=5 {~ms}^{-1}\end{aligned}\)
363493
Assertion : Even though net external force on a body is zero, momentum need not be conserved. Reason : The internal interaction between particles of a body cancels out momentum of each other.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Both Assertion and reason are incorrect.
Explanation:
Since \({\vec F_{ext{\rm{ }}}} = 0 \Rightarrow \frac{{d\vec p}}{{dt}} = 0\) \(\Rightarrow\) momentum gets conserved. Also the internal interactions involve internal forces. Internal forces cancel each other not the linear momenta.
AIIMS - 2019
PHXI05:LAWS OF MOTION
363494
Statement A : A body is momentarily at rest but still some force is acting on it at that time. Statement B : When a force acts on a body, it may not have some acceleration.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
A stationary body \((v = 0)\) may still have some acceleration, e.g. when a body is thrown in upward direction, it comes to rest at highest position, but at that time, it still have acceleration equal to acceleration due to gravity \(g\). Hence, gravitational force is acting at highest position and when a force acts on a body, then its accelerates. Therefore, Statement A is correct but Statement B is incorrect.
PHXI05:LAWS OF MOTION
363495
A gas molecule of mass \(m\) strikes the wall of the container with a speed \(v\) at an angle \(\theta \) with the normal to the wall at the point of collision. The impluse of the gas molecule has a magnitude
1 \(2\,mV\cos \theta \)
2 \(3\,mV\)
3 \({\rm{Zero}}\)
4 \(mV\)
Explanation:
From adjoining figure the component of momentum along \(x\)-axis (parallel to the wall of container) remains unchanged even after the collision. \(\therefore \) Impluse \(=\) change in momentum of gas molecule along \(y\)-axis, \( = 2mv\cos \theta \)
PHXI05:LAWS OF MOTION
363496
A body of mass \(4\,kg\) starts from rest, and a force is acting on the body, as shown in the figure. What is the speed of the body after \({4 s}\) ?
1 \({15 {~m} / {s}}\)
2 \({20 {~m} / {s}}\)
3 \({5 {~m} / {s}}\)
4 \({25 {~m} / {s}}\)
Explanation:
Area under graph, \(\begin{aligned}\Delta p & =\dfrac{1}{2} \times F \times t \\& =\dfrac{1}{2} \times 4 \times 10=20 {~kg} {~m} {~s}^{-1} \\\Delta p & =(m v-m u) \Rightarrow 20=m(v-u) \\20 & =4(v-0) v=5 {~ms}^{-1}\end{aligned}\)
