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dem82 [27]
4 years ago
12

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
Physics
1 answer:
Zolol [24]4 years ago
8 0

As per Newton's II law we can say

Force applied is rate of change of momentum

F = \frac{dP}{dt}

here we have

dP = P_f - P_i

dP = mv_f - mv_i

given that

m = 0.25 kg

v_f = 40 m/s

v_i = 0

F = 460 N

now by above equation

460 = \frac{0.25(40 - 0)}{t}

t = \frac{10}{460} = 0.022 s

so it required 0.022 s of contact time

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