Question

Two identical balls 'A' and 'B' are moving with the same velocity. If the velocity of 'A' is reduced to half and of 'B' to zero, then the rise in temperature of 'A' to that is reduced to half and of 'B' to zero, then the rise in temperature of 'A' to that of 'B' is
A) $3:4$
B) $4:1$
C) $2:1$
D) $1:1$

Answer 1

Hint: When the balls slow down, the loss in kinetic energy will be converted to heat energy. This heat energy will increase the temperature of the balls and the balls being identical will have the same specific heat capacity.

Formula used:
In this question, we will use the following formula
-$Q = mc\Delta T$ where $Q$ is the energy needed to increase the temperature of an object by $\Delta T$ temperature difference and $C$ is the specific heat capacity
- Kinetic energy of a ball: $K = \dfrac{1}{2}m{v^2}$ where \[m\] is the mass of the ball and $v$ is its velocity.

Complete step by step answer:
We’ve been told that two balls that are moving with the same velocity slow down and consequently both their temperature rises and we want to find the ratio of their temperature rise.
We will assume that the difference in kinetic energy will directly increase the temperature for both the balls, so we can write
$\Delta K = Q$
Since $\Delta K = \dfrac{1}{2}m{v_{final}}^2 - \dfrac{1}{2}m{v_{initial}}^2 = \dfrac{1}{2}m({v_{final}}^2 - {v_{initial}}^2)$, we can write
$\dfrac{1}{2}m({v_{final}}^2 - {v_{initial}}^2) = mc\Delta T$
Taking the ratio of kinetic energies for both the balls, we can write
$\dfrac{{\dfrac{1}{2}m({v_{1,final}}^2 - {v_{1,initial}}^2)}}{{\dfrac{1}{2}m({v_{2,final}}^2 - {v_{2,initial}}^2)}} = \dfrac{{mc\Delta {T_1}}}{{mc\Delta {T_2}}}$
Now, the velocity of 'A' is reduced to half and of 'B' to zero and the initial velocities of both the balls are the same which is denoted by $v$, so we can write
$\dfrac{{\dfrac{1}{2}m({v_{1,final}}^2 - {v_{1,initial}}^2)}}{{\dfrac{1}{2}m({v_{2,final}}^2 - {v_{2,initial}}^2)}} = \dfrac{{mc\Delta {T_1}}}{{mc\Delta {T_2}}}$
$\dfrac{{\dfrac{1}{2}\left[ {m\left\{ {{{\left( {\dfrac{v}{2}} \right)}^2} - {v^2}} \right\}} \right]}}{{\dfrac{1}{2}m(0 - {v^2})}} = \dfrac{{mc\Delta {T_1}}}{{mc\Delta {T_2}}}$
Which gives us
$\dfrac{3}{4} = \dfrac{{\Delta {T_1}}}{{\Delta {T_2}}}$

Hence the correct choice is option (A).

Note: Here we have assumed that the loss in kinetic energy will all be converted into changing the temperature of the balls however in reality there are energy losses associated with air drag resistance and other friction forces. As a result, balls with different velocities will experience different energy losses, and different energy rises than in the ideal case.