Answer to Question #148153 in Physics for Jesca

Question

A 55.0 kg player vaults to a height of 6.5 m before dropping to a think foam mattress that was placed to cushion her fall. Find (a) the momentum of the player just before landing and (b) the force exerted by the mattress if the player was stopped in 0.65s.

Answer 1

a) Let us use the conservation of energy law: the kinetic energy of the player just before landing$K=\dfrac{p^2}{2m}$ is equal to his potential energy in the highest point $U=mgh$ . Thus, obtain:

$$\dfrac{p^2}{2m} = mgh$$

where $p$ is the  the momentum of the player just before landing, $m = 55kg$ is his mass, $g = 9.81m/s^2$ is the gravitational acceleration, and $h = 6.5m$ is the maximal height.

From this equation find $p$:

$$p = m\sqrt{2gh}\\ p = 55\cdot \sqrt{2\cdot 9.81\cdot 6.5} \approx 621.11\space kg\cdot m/s$$

b) According to Newton's second law, the magnitude of force exerted by the mattress is equal to:

$$F = \left|\dfrac{\Delta p}{\Delta t}\right|$$

where $\Delta p = 0-p = 0-621.11 = -621.11\space kg\cdot m/s$ is the change is momentum (from $621.11 \space kg\cdot m/s$ to $0\space kg\cdot m/s$) happened in time $\Delta t = 0.65s$. Thus, obtain:

$$F = \left|\dfrac{-621.11 \space kg\cdot m/s}{0.65s}\right| \approx 955.55\space N$$

Answer. a) $621.11 \space kg\cdot m/s$, b) $955.55\space N$.