Question #8dc0a

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Answer 1 · 2017-02-12T09:35:37

Given

$Mass of the ball m = 300 g = 0.3 k g$ $"Mass of the ball "m=300g=0.3kg$

$Initial velocity of the ball u = - 8 m / s$ $"Initial velocity of the ball "u=-8m"/"s$

$Final velocity of the ball v = + 6 m / s$ $"Final velocity of the ball "v=+6m"/"s$

$Force applied on the ball by the wall F = 2000 N$ $"Force applied on the ball by the wall "F=2000N$

$Let the time of ball-wall contact be = t s$ $"Let the time of ball-wall contact be "=t s$

So by Newton's law we have

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

$\Rightarrow 2000 = \frac{0.3 \times (6 - (- 8))}{t}$ $=>2000=(0.3xx(6-(-8)))/t$

$\Rightarrow t = \frac{4.2}{2000} = 2.1 \times 10^{- 3} s = 2.1 m s$ $=>t=4.2/2000=2.1xx10^(-3)s=2.1ms$