Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

The question has Statement I and Statement II. Of the four choices given after the Statements, choose the one that best describes the two statements.
Statement – I: A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as f $(\dfrac{1}{2}m{{v}^{2}})$ then $f=\left( \dfrac{m}{M+m} \right)$ .
Statement – II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.
A: Statement I is true, Statement II is true, Statement II is not a correct explanation of Statement I.
B: Statement I is true, Statement II is false.
C: Statement I is false, Statement II is true.
D: Statement I is true, Statement II is true, Statement II is a correct explanation of Statement I.

seo-qna
Last updated date: 16th Jun 2024
Total views: 402.9k
Views today: 8.02k
Answer
VerifiedVerified
402.9k+ views
Hint: The maximum energy loss is the difference in kinetic energy before collision and kinetic energy after collision. When two objects collide the total momentum before collision is equal to the momentum after collision in the absence of external forces. That is the law of conservation of momentum.
Formula used:
Kinetic energy, $K.E=\dfrac{{{p}^{2}}}{2m}$

Complete answer:
Let us consider two particles of masses ${{m}_{1}}$ and ${{m}_{2}}$ moving with velocities ${{u}_{1}}$ and ${{u}_{2}}$ respectively before collision. If there velocities after collision are ${{v}_{1}}$ and ${{v}_{2}}$, then according to conservation of momentum we have,
${{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$
Here, the initial and final positions are widely separated so that the interaction forces between the particles becomes effectively zero. Hence the potential energy before and after remains the same. If the collision is perfectly elastic, the total kinetic energy of the particles is not changed by the collision.
$\dfrac{1}{2}{{m}_{1}}u_{1}^{2}+\dfrac{1}{2}{{m}_{2}}u_{2}^{2}=\dfrac{1}{2}{{m}_{1}}v_{1}^{2}+\dfrac{1}{2}{{m}_{2}}v_{2}^{2}$
According to the kinetic theory of gases, such elastic collision occurs between the molecules of a gas. This type of collision mostly takes place between atoms, electrons and protons.
In case of inelastic collision, a part of kinetic energy is converted to some other forms. This energy appears in the form of thermal energy in macroscopic particles.
Let p be the momentum before and after collision.
We know that,
Kinetic energy$K.E=\dfrac{{{p}^{2}}}{2m}$
Maximum energy loss $=\dfrac{{{p}^{2}}}{2m}-\dfrac{{{p}^{2}}}{2(m+M)}$
               $=\dfrac{{{p}^{2}}}{2m}\left[ 1-\dfrac{m}{m+M} \right]$
                $=\dfrac{1}{2}m{{v}^{2}}\left[ \dfrac{M}{m+M} \right]$
Hence statement I is false.
In inelastic collision loss of energy is maximum.
Thus statement II is true.

Thus option (C) is correct.

Note:
Maximum energy loss occurs when the particles strike together as a result of collision. The collision imparts force only along the line of collision, the collision does not change the velocities that are tangent to the point of collision. The velocities along the line of collision can be used as the same one dimensional equation.