Question

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by 0.50 m , the mass that moves downward is 70 kg , and the collision on the floor lasts 0.082 s. What are the magnitudes of the (a) Impulse and (b) Average force acting on the victim from the floor during the collision?

Short answer

1. The impulse acting on the victim,$\mathrm{J}=2.2\times {10}^2\mathrm{N}·\mathrm{s}$.
2. The magnitude of the average force on the victim from the floor, ${\mathrm{F}}_{\mathrm{avg}}=2.7\times {10}^3\mathrm{N}$.

Step-by-step solution

Understanding the given information

The mass of the victim,$\mathrm{m}=70\mathrm{kg}$.

The height of the victim from the floor,$\mathrm{h}=0.50\mathrm{m}$.

The time of contact, $\mathrm{t}=0.082\mathrm{s}$.

Concept and formula used in the given question

We can use the equation of impulse related to the change in momentum to calculate the impulse on the ball. The magnitude of the average force can be calculated using an equation relating impulse, force, and time which are given as.

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ ∆\mathrm{P}=\mathrm{m}\times \left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ \mathrm{J}={\mathrm{F}}_{\mathrm{avg}}\times \mathrm{t} \end{gathered}$$

(a) Calculation for the magnitude of the impulse

Let us calculate the velocity ofthevictim just before the collision.

The velocity of free fall from height ‘h’ is

$$\begin{gathered} \mathrm{v}={\mathrm{v}}_1 \\ =\sqrt{2\mathrm{gh}} \\ =\sqrt{2\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 0.50\mathrm{m}} \\ =3.13\mathrm{m}/\mathrm{s} \end{gathered}$$

Now,

The equation of the impulse-momentum theorem is

$$\begin{gathered} \mathrm{J}=∆\mathrm{P} \\ =\mathrm{m}\times \left({\mathrm{v}}_2-{\mathrm{v}}_1\right) \\ =70\mathrm{kg}\times \left(0\mathrm{m}/\mathrm{s}-\left(-3.13\mathrm{m}/\mathrm{s}\right)\right) \\ =2.2\times {10}^2\mathrm{N}·\mathrm{s} \end{gathered}$$

(b) Calculation for the magnitude of average force acting on the victim from the floor during the collision

We have,

$\mathrm{J}={\mathrm{F}}_{\mathrm{avg}}\times \mathrm{t}$

So,

$$\begin{gathered} {\mathrm{F}}_{\mathrm{avg}}=\frac{\mathrm{J}}{\mathrm{t}} \\ =\frac{2.2\times {10}^2\mathrm{N}·\mathrm{s}}{0.082\mathrm{s}} \\ =2.7\times {10}^3\mathrm{N} \end{gathered}$$