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{ = }(0.25*(40-0))/(t) \\ 460t\text{ = 0.25(40)} \\ 460t\text{ = }10 \\ t\text{ = }(10)/(460) \\ t\text{ = }0.0217s \end{gathered}`

Therefore, the time of impact = 0.0217 seconds