A bullet of mass 50 g is fired with a speed 100 m/s on a plywood and emerges with 40 m/s. The percentage loss of kinetic energy is :
- 0.32
- 0.44
- 0.16
- 0.84
The Correct Option is D
Approach Solution - 1
The initial kinetic energy of the bullet is:
\[ K_i = \frac{1}{2} m (100)^2 \]
The final kinetic energy of the bullet is:
\[ K_f = \frac{1}{2} m (40)^2 \]
The percentage loss in kinetic energy is given by:
\[ \% \text{loss} = \frac{|K_f - K_i|}{K_i} \times 100 \]
Substituting the expressions for \( K_i \) and \( K_f \):
\[ \% \text{loss} = \frac{\left| \frac{1}{2} m (40)^2 - \frac{1}{2} m (100)^2 \right|}{\frac{1}{2} m (100)^2} \times 100 \]
Simplify:
\[ \% \text{loss} = \frac{|1600 - 10000|}{10000} \times 100 \]
\[ \% \text{loss} = \frac{8400}{10000} \times 100 = 84\% \]
Final Answer: 84% (Option 4)
Approach Solution -2
The problem asks for the percentage loss of kinetic energy when a bullet of a given mass and initial speed passes through plywood and emerges with a reduced speed.
Concept Used:
1. Kinetic Energy (KE): The kinetic energy of an object of mass \( m \) moving with a speed \( v \) is given by the formula:
\[ KE = \frac{1}{2}mv^2 \]2. Percentage Loss: The percentage loss in any quantity is calculated as the ratio of the change (loss) in the quantity to its initial value, multiplied by 100.
\[ \text{Percentage Loss} = \frac{\text{Initial Value} - \text{Final Value}}{\text{Initial Value}} \times 100\% \]In this case, the percentage loss in kinetic energy is:
\[ \% \text{ Loss in KE} = \frac{KE_{\text{initial}} - KE_{\text{final}}}{KE_{\text{initial}}} \times 100\% \]Step-by-Step Solution:
Step 1: List the given parameters and convert them to SI units.
Given:
- Mass of the bullet, \( m = 50 \, \text{g} = 50 \times 10^{-3} \, \text{kg} = 0.05 \, \text{kg} \)
- Initial speed of the bullet, \( v_i = 100 \, \text{m/s} \)
- Final speed of the bullet, \( v_f = 40 \, \text{m/s} \)
Step 2: Calculate the initial kinetic energy (\( KE_i \)) of the bullet.
\[ KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(0.05 \, \text{kg})(100 \, \text{m/s})^2 \] \[ KE_i = \frac{1}{2}(0.05)(10000) \, \text{J} = 0.05 \times 5000 \, \text{J} = 250 \, \text{J} \]Step 3: Calculate the final kinetic energy (\( KE_f \)) of the bullet after it emerges from the plywood.
\[ KE_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(0.05 \, \text{kg})(40 \, \text{m/s})^2 \] \[ KE_f = \frac{1}{2}(0.05)(1600) \, \text{J} = 0.05 \times 800 \, \text{J} = 40 \, \text{J} \]Step 4: Calculate the loss in kinetic energy (\( \Delta KE \)).
\[ \Delta KE = KE_i - KE_f = 250 \, \text{J} - 40 \, \text{J} = 210 \, \text{J} \]Final Computation & Result:
Step 5: Calculate the percentage loss in kinetic energy.
\[ \% \text{ Loss in KE} = \frac{\Delta KE}{KE_i} \times 100\% \] \[ \% \text{ Loss in KE} = \frac{210 \, \text{J}}{250 \, \text{J}} \times 100\% \] \[ \% \text{ Loss in KE} = \frac{21}{25} \times 100\% = 21 \times 4\% = 84\% \]Alternative Method:
The percentage loss can also be calculated without finding the absolute KE values, which simplifies the calculation:
\[ \% \text{ Loss in KE} = \frac{\frac{1}{2}mv_i^2 - \frac{1}{2}mv_f^2}{\frac{1}{2}mv_i^2} \times 100\% = \frac{v_i^2 - v_f^2}{v_i^2} \times 100\% \] \[ \% \text{ Loss in KE} = \left(1 - \frac{v_f^2}{v_i^2}\right) \times 100\% = \left(1 - \left(\frac{v_f}{v_i}\right)^2\right) \times 100\% \] \[ \% \text{ Loss in KE} = \left(1 - \left(\frac{40}{100}\right)^2\right) \times 100\% = \left(1 - (0.4)^2\right) \times 100\% \] \[ \% \text{ Loss in KE} = (1 - 0.16) \times 100\% = 0.84 \times 100\% = 84\% \]Thus, the percentage loss of kinetic energy is 84%.
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