haleygrobertson6548 haleygrobertson6548
  • 18-10-2018
  • Physics
contestada

A .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

Respuesta :

aristocles aristocles
  • 28-10-2018

As per Newton's II law we can say

Force applied is rate of change of momentum

[tex]F = \frac{dP}{dt}[/tex]

here we have

[tex]dP = P_f - P_i[/tex]

[tex]dP = mv_f - mv_i[/tex]

given that

m = 0.25 kg

[tex]v_f = 40 m/s[/tex]

[tex]v_i = 0[/tex]

F = 460 N

now by above equation

[tex]460 = \frac{0.25(40 - 0)}{t}[/tex]

[tex]t = \frac{10}{460} = 0.022 s[/tex]

so it required 0.022 s of contact time

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