QJEEM24Ph75d9: JEE Main 2024 Physics - Class 11 JEEM24Ph75d9 Solution: A bullet of mass is fired with a speed on a plywood and emerges …

To find the percentage loss of kinetic energy of the bullet, we first calculate the initial kinetic energy before the bullet hits the plywood and the final kinetic energy after it emerges. The formula for kinetic energy (KE) is given by:

$KE = \frac{1}{2} mv^2$

where $m$ is the mass of the object and $v$ is its velocity.

Let's calculate the initial and final kinetic energies.

Initial Kinetic Energy:

$KE_{\text{initial}} = \frac{1}{2} \times 50 \times (100)^2 = \frac{1}{2} \times 50 \times 10000 = 25 \times 10000 = 250000 \, \text{g.m}^2/\text{s}^2$

Note: To keep units consistent, we used grams and meters per second. We can also convert the mass to kilograms (by dividing by 1000) which would result in the energy being calculated in Joules, but for the purpose of finding the percentage change, the form of units does not matter as long as they are consistent, since it will be a ratio.

Final Kinetic Energy:

$KE_{\text{final}} = \frac{1}{2} \times 50 \times (40)^2 = \frac{1}{2} \times 50 \times 1600 = 25 \times 1600 = 40000 \, \text{g.m}^2/\text{s}^2$

The loss of kinetic energy is then:

$\Delta KE = KE_{\text{initial}} - KE_{\text{final}} = 250000 - 40000 = 210000 \, \text{g.m}^2/\text{s}^2$

Finally, the percentage loss of kinetic energy can be calculated using the formula:

$\text{Percentage loss of KE} = \left( \frac{\Delta KE}{KE_{\text{initial}}} \right) \times 100\%$

$\text{Percentage loss of KE} = \left( \frac{210000}{250000} \right) \times 100\% = 0.84 \times 100\% = 84\%$

Thus, the percentage loss of kinetic energy is 84%, which corresponds to Option C.