
Consider the following two statements.
(A) Linear momentum of a system of particles is zero.
(B) Kinetic energy of a system of particles is zero.
A. A does not imply B and B does not imply A.
B. A implies B but B does not imply A.
C. A does not imply B but B implies A.
D. A implies B and B implies A.
Answer
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Hint: The vector quantity linear momentum is defined as the product of an object's velocity, v, and mass, m. Kinetic energy is the energy of action, which can be observed as an item, particle, or group of particles moving.
Formula Used:
\[p = mv\] and \[KE = \dfrac{{pv}}{2}\]
where KE is the kinetic energy, p is the linear momentum, m is the mass and v is the velocity.
Complete step by step solution:
For the given statements,
1. Linear momentum of a system of particles is zero.
2. Kinetic energy of a system of particles is zero.
We have to find out how one statement affects the other statements. We will make two cases in which we will keep one of the statements true and try to prove the other statement.
Case 1: Let us assume that there are two particles of equal mass and equal velocity, thus the equal momentum \[{p_1} = {p_2}\], going in the opposite direction. So, as two particles having equal momentum are going in the opposite direction, we can say that the momentum of any one particle with respect to the other will be zero,
\[|{p_1}| = |{p_2}| \\
\Rightarrow |{p_1}| \equiv 0 \\
\Rightarrow |{p_2}| \equiv 0 \]
So, the kinetic energy of the system will be,
\[KE = \dfrac{{{p_1}{v_1}}}{2} + \dfrac{{{p_2}{v_2}}}{2} \\
\Rightarrow KE = {p_1}{v_1} \\ \]
As only the magnitude of the system of momentum is zero, it does not mean that the particles are in a state of rest. Even if the system of linear momentum is zero, the particles are still producing the kinetic energy, one particle going in one direction and the other particle going in the opposite direction, and so the kinetic energy of both the particles will be added giving the kinetic energy of the system. As the momentum and velocity of both the particles are the same, the total kinetic energy will be twice that of either one of their kinetic energy. So, this proves that statement A does not imply statement B.
Case 2: Now, if we consider that the kinetic energy of a system of particles is zero, we can say that it will only be zero if both the particles are in a state of rest. If even one of the particles moves from its position, it will produce kinetic energy. So, as both the particles are in a state of rest, we can also say that the linear momentum of a system of particles is also zero. So, this proves that statement B implies statement A.
So, option C is the required solution.
Note: The meaning of A implies B is that if statement A is true then statement B must be true. The meaning of A does not imply B means that if statement A is true then statement B can or cannot be true, depending upon the situation.
Formula Used:
\[p = mv\] and \[KE = \dfrac{{pv}}{2}\]
where KE is the kinetic energy, p is the linear momentum, m is the mass and v is the velocity.
Complete step by step solution:
For the given statements,
1. Linear momentum of a system of particles is zero.
2. Kinetic energy of a system of particles is zero.
We have to find out how one statement affects the other statements. We will make two cases in which we will keep one of the statements true and try to prove the other statement.
Case 1: Let us assume that there are two particles of equal mass and equal velocity, thus the equal momentum \[{p_1} = {p_2}\], going in the opposite direction. So, as two particles having equal momentum are going in the opposite direction, we can say that the momentum of any one particle with respect to the other will be zero,
\[|{p_1}| = |{p_2}| \\
\Rightarrow |{p_1}| \equiv 0 \\
\Rightarrow |{p_2}| \equiv 0 \]
So, the kinetic energy of the system will be,
\[KE = \dfrac{{{p_1}{v_1}}}{2} + \dfrac{{{p_2}{v_2}}}{2} \\
\Rightarrow KE = {p_1}{v_1} \\ \]
As only the magnitude of the system of momentum is zero, it does not mean that the particles are in a state of rest. Even if the system of linear momentum is zero, the particles are still producing the kinetic energy, one particle going in one direction and the other particle going in the opposite direction, and so the kinetic energy of both the particles will be added giving the kinetic energy of the system. As the momentum and velocity of both the particles are the same, the total kinetic energy will be twice that of either one of their kinetic energy. So, this proves that statement A does not imply statement B.
Case 2: Now, if we consider that the kinetic energy of a system of particles is zero, we can say that it will only be zero if both the particles are in a state of rest. If even one of the particles moves from its position, it will produce kinetic energy. So, as both the particles are in a state of rest, we can also say that the linear momentum of a system of particles is also zero. So, this proves that statement B implies statement A.
So, option C is the required solution.
Note: The meaning of A implies B is that if statement A is true then statement B must be true. The meaning of A does not imply B means that if statement A is true then statement B can or cannot be true, depending upon the situation.
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