Physics Test 93...

Two particles are in SHM with same amplitude A and same angular frequency ω. At time t = 0, one is at x = +A/2and other is at x = –A/2. Both are moving in same direction.

Nuclear reactions, obey the usual conservation laws for energy (including rest-mass energy), momentum, angular momentum (including spin), electric charge as well as mass number.
Suppose that the mass of the projectile is m₁, the mass of the target nucleus is m₂, and the masses of the two nuclei produced in the reaction are m'₁ and m'₂. We designate the change of rest-mass energy by Q as below : 

Q = (m₁ + m₂ – m'₁ – m'₂)c²

The total energy of each particle of nucleus is the sum of its kinetic energy and its rest-mass energy. Thus, the law of conservation of energy for the reaction is
K + Q = K’
Reactions with a positive value of Q convert some of the initial rest-mass energy into kinetic energy; such reactions are called exoergic. Reactions with a negative value of Q convert some of the initial kinetic energy into rest-mass energy; these reactions are called endoergic. An endoergic reaction is not viable unless the initial kinetic energy is sufficiently large. The initial kinetic energy must provide not only the increase in rest-mass energy but also the energy for the (constant) motion of the center of mass. According to equation, in a one-dimensional collision, the available portion of the initial kinetic energy (the “internal” kinetic energy) is 0.5 µ (v₁ – v₂)² or, since the target is at rest, so 0.5 µv₁², where µ = m₁m₂/ (m₁ + m₂). Thus, the available kinetic energy is smaller than the actual kinetic energy of the projectile by a factor of m₂/(m₁ + m₂). For the reaction to proceed, this available kinetic energy must be at least as large as –Q so 0.5 µv₁² ≥ –Q or 0.5 m₁v₁² ≥ (m₁ + m₂)(– Q)/m₂ 

Nuclear reactions, obey the usual conservation laws for energy (including rest-mass energy), momentum, angular momentum (including spin), electric charge as well as mass number.

Suppose that the mass of the projectile is m₁, the mass of the target nucleus is m₂, and the masses of the two nuclei produced in the reaction are m'₁ and m'₂. We designate the change of rest-mass energy by Q as below : 

Q = (m₁ + m₂ – m'₁ – m'₂)c²

The total energy of each particle of nucleus is the sum of its kinetic energy and its rest-mass energy. Thus, the law of conservation of energy for the reaction is

K + Q = K’
Reactions with a positive value of Q convert some of the initial rest-mass energy into kinetic energy; such reactions are called exoergic. Reactions with a negative value of Q convert some of the initial kinetic energy into rest-mass energy; these reactions are called endoergic. An endoergic reaction is not viable unless the initial kinetic energy is sufficiently large. The initial kinetic energy must provide not only the increase in rest-mass energy but also the energy for the (constant) motion of the center of mass. According to equation, in a one-dimensional collision, the available portion of the initial kinetic energy (the “internal” kinetic energy) is 0.5 µ (v₁ – v₂)² or, since the target is at rest, so 0.5 µv₁², where µ = m₁m₂/ (m₁ + m₂). Thus, the available kinetic energy is smaller than the actual kinetic energy of the projectile by a factor of m₂/(m₁ + m₂). For the reaction to proceed, this available kinetic energy must be at least as large as –Q so 0.5 µv₁² ≥ –Q or 0.5 m₁v₁² ≥ (m₁ + m₂)(– Q)/m₂ 

Mark the INCORRECT statement(s) :