Question

A 0.25 kg ball initially at rest is hit with a 460 N impact. What is the time of impact for this event if the ball ends up moving 40 m/s?

Answer 1

Answer:

Time of impact = 0.0217 seconds

Explanations:

The mass of the ball, m = 0.25 kg

The ball is initially at rest

That is, the initial velcity, u = 0 m/s

The impact force, F = 460 N

The final velocity, v = 40 m/s

The equation below should be solved to get the time of impact

$F\text{ = }\frac{m(v\text{ - u)}}{t}$

Substituting the values of F, m, v, and u into the formula above

$\begin{gathered} 460\text{ = }\frac{0.25\times(40-0)}{t} \\ 460t\text{ = 0.25(40)} \\ 460t\text{ = }10 \\ t\text{ = }\frac{10}{460} \\ t\text{ = }0.0217s \end{gathered}$

Therefore, the time of impact = 0.0217 seconds