Q18DQ - University Physics with Modern Physics

Question

“It’s not the fall that hurts you; it’s the sudden stop at the bottom.” Translate this saying into the language of Newton’s laws of motion.

Step-by-Step Solution

Answer

It’s not the $\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}$that hurts you; it’s the sudden change in the momentum i.e.$\mathit{F}\mathbf{=}\mathbf{-}\frac{\mathbf{m}\mathbf{u}}{\mathbf{t}}$at the bottom.

Step 1: Newton’s Second law

A body accelerates when a net force acts on it and travels in the same direction as the net force. The product of mass and acceleration vector is equal to the force vector.

$\sum \overrightarrow{F}=m\overrightarrow{a}$

Step 2: Kinematic equation in terms of acceleration, velocity, and initial velocity.

For constant acceleration, the final velocity v is related to the initial velocity u by the expression given below,

$\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}$

The statement means that the high final velocity vdoes not hurt you in a fall, but the sudden change in velocity to zero hurts. The force acted on the body at the end of the fall will depend on the de-acceleration and the time it took to stop the fall completely.

$$\begin{gathered} F=ma \\ =m\left(\frac{v-u}{t}\right) \\ =m\left(\frac{0-u}{t}\right) \\ =-\frac{mu}{t} \end{gathered}$$

Hence, it can be said that It’s not the $\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}$ that hurts you; it’s the sudden change in the momentum, i.e. $\mathit{F}\mathbf{=}\mathbf{-}\frac{\mathbf{m}\mathbf{u}}{\mathbf{t}}$at the bottom.