If two balls of same metal weighing 5 gm and 10 gm strike with a target with the same velocity. The heat energy so developed is used for raising their temperature alone, then the temperature will be higher

Options

A. For bigger ball
B. For smaller ball
C. Equal for both the balls
D. None is correct from the above three

Explanation

To solve the problem, we need to analyze the heat energy developed when the two balls strike a target and how that energy affects their temperatures.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Mass of ball 1 (m1) = 5 gm = 0.005 kg
- Mass of ball 2 (m2) = 10 gm = 0.010 kg
- Both balls are made of the same metal, so they have the same specific heat capacity (C).
- Both balls strike the target with the same velocity (v).

2. **Calculate the Kinetic Energy of Each Ball**:
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- For ball 1 (5 gm):
\[ KE_1 = \frac{1}{2} m_1 v^2 = \frac{1}{2} (0.005) v^2 = 0.0025 v^2 \]
- For ball 2 (10 gm):
\[ KE_2 = \frac{1}{2} m_2 v^2 = \frac{1}{2} (0.010) v^2 = 0.005 v^2 \]

3. **Heat Energy Developed**:
The kinetic energy of each ball is converted into heat energy when they strike the target. Thus:
- Heat energy developed by ball 1 (Q1) = KE1 = \(0.0025 v^2\)
- Heat energy developed by ball 2 (Q2) = KE2 = \(0.005 v^2\)

4. **Relate Heat Energy to Temperature Change**:
The heat energy absorbed by an object is related to its mass, specific heat capacity, and change in temperature (ΔT) by the formula:
\[ Q = mc\Delta T \]
Rearranging gives us:
\[ \Delta T = \frac{Q}{mc} \]

5. **Calculate the Temperature Change for Each Ball**:
- For ball 1 (5 gm):
\[ \Delta T_1 = \frac{Q_1}{m_1 C} = \frac{0.0025 v^2}{0.005 C} = \frac{0.5 v^2}{C} \]
- For ball 2 (10 gm):
\[ \Delta T_2 = \frac{Q_2}{m_2 C} = \frac{0.005 v^2}{0.010 C} = \frac{0.5 v^2}{C} \]

6. **Compare the Temperature Changes**:
From the calculations:
- \(\Delta T_1 = \frac{0.5 v^2}{C}\)
- \(\Delta T_2 = \frac{0.5 v^2}{C}\)

Since both temperature changes (\(\Delta T_1\) and \(\Delta T_2\)) are equal, we conclude that the temperature increase for both balls is the same.

### Conclusion:
The temperature increase will be the same for both balls. Therefore, neither ball will have a higher temperature; they will both reach the same final temperature after the heat energy is absorbed.

### Final Answer:
The temperature will be the same for both balls.

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