Question

7. A ball is dropped from a height of 4.0 m. Just before it hits the ground it's momentum is

Answer 1

Answer:

Approximately $1\; \rm kg$. (Assumption: $g = 9.8\; \rm m \cdot s^{-2}$.)

Explanation:

Let the mass of this ball be $x\; \rm kg$.

Initial gravitational potential energy of this ball: $m \cdot g \cdot h \approx (39.2\, x) \; \rm J$.

Just before the ball hits the ground, all $(39.2\, x)\; \rm J$ of gravitational potential energy would have been converted to kinetic energy. Calculate the velocity of the ball at that moment:

$\begin{aligned} \text{kinetic energy} = \frac{1}{2}\, m \cdot v^{2} \end{aligned}$.

Therefore, right before hitting the ground, the velocity of the ball would be:

$\begin{aligned} v&= \sqrt{\frac{2\, (\text{kinetic energy})}{m}} \\ &= \sqrt{\frac{2 \times (39.2\, x)}{x}}\\ & \approx \sqrt{2 \times 39.2} \approx 8.85\; \rm m \cdot s^{-1}\end{aligned}$.

The momentum $p$ of an object of mass $m$ and velocity $v$ would be $p = m \cdot v$. Rewrite this equation to find an expression for mass $m\!$ given velocity $v\!$ and momentum $p\!$:

$\displaystyle m = \frac{p}{v}$.

Right before collision, the momentum of this ball is $p = 8.85\; \rm kg \cdot m \cdot s^{-1}$ while its velocity is $v \approx 8.85\; \rm m \cdot s^{-1}$. Therefore, the mass of this ball would be:

$\begin{aligned}m &= \frac{p(\text{right before landing})}{v(\text{right before landing})} \\ &\approx \frac{8.85\; \rm kg \cdot m \cdot s^{-1}}{8.85\; \rm m \cdot s^{-1}} \approx 1\; \rm kg\end{aligned}$.