Why is there a 1/2 in the formula of kinetic energy?

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Aadit Singh Uppal | Contributor-Level 10

6 months ago

The 1/2 is a result of mathematical calculation, which occurs when we integrate? vdv in the formula of work done according to Newton's second law of motion. Without this, the final result will turn out to be twice of the actual value.

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A smooth spherical ball strikes a smooth horizontal floor at an angle $θ = 45^{?}$ . The coefficient of restitution between the ball and the floor is $e = \frac{1}{2}$ . The fraction of its kinetic energy lost in collision is: (there is no rotational kinetic energy before or after collision)


Raj Pandey

Vertical component of velocity just after collision $= \frac{u}{2 \sqrt[]{2}}$

$k_{i} = \frac{m u^{2}}{2} k_{f} = \frac{1}{2} m \frac{u^{2}}{2} + \frac{1}{2} m \frac{u^{2}}{8} = \frac{5 m u^{2}}{16}$ $\frac{}{}$

Fraction $= \frac{k_{i} - k_{f}}{k_{i}} = 1 - \frac{\frac{5 m u^{2}}{16}}{\frac{m u^{2}}{2}} = 1 - \frac{5}{8} = \frac{3}{8}$

As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed of the particle at B is x m/s. (Take g = 10 m/s²) The value of 'x' to the nearest integer is----------.

Raj Pandey

Using Conservation of Mechanical Energy at point-A and at point-B, we can write
K_B = U_A - U_B [Since K_A = 0]
⇒ (1/2)mv_B² = mg (h_A - h_B)
⇒ v_B = √ (2 × 10 × (10 - 5) = 10m/s

A body $A$ , of mass $m = 0.1 k g$ has an initial velocity of $3 \hat{i} {m s}^{- 1}$ $^{}$ . It collides elastically with another body, $B$ of the same mass which has an initial velocity of $5 \hat{j} {m s}^{- 1}$ $^{}$ . After collision, A moves with a velocity $\overset{?}{v} = 4 (\hat{i} + \hat{j})$ . The energy of $B$ after collision is written as $\frac{}{}$ $\frac{x}{10} J$ . The value of $x$ is

Raj Pandey

For elastic collision ${K E}_{i} = {K E}_{r}$

$\frac{1}{2} m \times 25 + \frac{1}{2} \times m \times 9 = \frac{1}{2} m \times 32 + \frac{1}{2} m v^{2}$

$34 = 32 + v^{2}$

$K E = \frac{1}{2} \times 0.1 \times 2 = 0.1 J = \frac{1}{10}$

$x = 1$

Can changing the direction of a body also affect it's kinetic energy?

Aadit Singh Uppal

No. Since kinetic energy is a scalar quantity, it only depends on speed of the body and not the direction. So if the direction of the body is changed but the speed remains unchanged, there won't be any effect on the kinetic energy. However, if changing the direction also changes the speed of the bod

Aadit Singh Uppal

If you look closely at the formula of kinetic energy (1/2*m*v^2), the velocity is squared which automatically gives a positive integer. And mass of the body can never be a negative value, which leads to the result being a positive integer.