Question

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

8.85 kg-m/s. What is the mass of the ball?

Answer 1

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

Step-by-step 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} = (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{(2 * (39.2\, x))/(x)}\\ & \approx √(2 * 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 = (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 (8.85\; \rm kg \cdot m \cdot s^(-1))/(8.85\; \rm m \cdot s^(-1)) \approx 1\; \rm kg\end{aligned}`.