If the velocity of a moving body is reduced to half of its initial value while mass remains the same, how does its kinetic energy change?

Introduction / Context:
Kinetic energy depends on both the mass of a body and the square of its velocity. This question tests conceptual understanding of how kinetic energy changes when the velocity of a body changes, especially when the velocity is scaled by a factor. Recognising the square relationship between speed and kinetic energy is important for reasoning about collisions, braking distances, and safety considerations in everyday life and engineering.

Given Data / Assumptions:
• The mass of the body remains constant. • The initial velocity is some value v. • The new velocity is v / 2, that is half the original. • We are asked how the kinetic energy changes relative to its initial value.

Concept / Approach:
The formula for kinetic energy is KE = (1 / 2) * m * v^2. If we denote the initial kinetic energy by KE_initial and the new kinetic energy by KE_new, we can write KE_initial = (1 / 2) * m * v^2 and KE_new = (1 / 2) * m * (v / 2)^2. By taking the ratio KE_new / KE_initial, we can see directly how kinetic energy scales when velocity is halved. This is a simple algebraic manipulation that clearly illustrates the square dependence on velocity.

Step-by-Step Solution:
Step 1: Write the initial kinetic energy KE_initial = (1 / 2) * m * v^2. Step 2: Write the new velocity as v / 2. Step 3: Compute the new kinetic energy KE_new = (1 / 2) * m * (v / 2)^2. Step 4: Simplify (v / 2)^2 = v^2 / 4. Step 5: Substitute into KE_new: KE_new = (1 / 2) * m * (v^2 / 4) = (1 / 2) * (1 / 4) * m * v^2 = (1 / 8) * m * v^2. Step 6: Compare KE_new with KE_initial = (1 / 2) * m * v^2. The ratio KE_new / KE_initial = (1 / 8) / (1 / 2) = (1 / 8) * 2 = 1 / 4. Step 7: Therefore, the new kinetic energy is one quarter of the original kinetic energy.

Verification / Alternative check:
We can verify with simple numbers. Assume m = 1 kg and initial velocity v = 4 m/s. Then KE_initial = (1 / 2) * 1 * 16 = 8 J. Halving the velocity gives v / 2 = 2 m/s. Then KE_new = (1 / 2) * 1 * 4 = 2 J. The ratio 2 / 8 is 1 / 4, confirming that the kinetic energy becomes one quarter when velocity is halved. This numerical example matches the algebraic result.

Why Other Options Are Wrong:
Option a (It becomes 4 times): This would correspond to doubling the velocity, not halving it, because kinetic energy is proportional to v^2. Option c (It is also doubled): There is no mathematical reason for doubling when velocity is halved. Option d (It becomes half): This would be true for a linear dependence on velocity, but kinetic energy depends on the square of velocity, so half speed gives one quarter energy. Option e (It remains unchanged): This would only be true if kinetic energy were independent of velocity, which is not correct.

Common Pitfalls:
Learners sometimes think in linear terms and assume that if velocity is halved, kinetic energy is also halved. This mistake comes from forgetting the square in the kinetic energy formula. Another error is manipulating fractions incorrectly when squaring v / 2. To avoid these problems, always write the expression for kinetic energy with v^2, do the algebra step by step, and if needed test with simple numeric values to see the effect.

Final Answer:
When the velocity is halved, the kinetic energy becomes 1/4 of its initial value.