A time-varying force acts on a ball of mass 1 | vedclass

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(B) The impulse imparted to the ball is equal to the area under the $F-t$ graph.Impulse = Area of the triangle = $\frac{1}{2} 	imes 	ext{base} 	ime

Vedclass · Accepted Answer

$210$

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(B) The impulse imparted to the ball is equal to the area under the $F-t$ graph.Impulse = Area of the triangle = $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$Impulse = $\frac{1}{2} 	imes (2 	imes 10^{-3} \ s) 	imes (20 	imes 10^3 \ N)$Impulse = $20 \ N \ s = 20 \ kg \ m \ s^{-1}$According to the impulse-momentum theorem,impulse is equal to the change in momentum $(\Delta p)$:$\Delta p = m v - m u = 	ext{Impulse}$Given: mass $m = 100 \ g = 0.1 \ kg$,initial velocity $u = 10 \ m \ s^{-1}$.$0.1 	imes v - 0.1 	imes 10 = 20$$0.1 	imes v - 1 = 20$$0.1 	imes v = 21$$v = \frac{21}{0.1} = 210 \ m \ s^{-1}$

$A$ time-varying force acts on a ball of mass $100 \ g$ for $2 \ ms$. The force versus time curve is shown below. If the initial speed of the ball is $10 \ m \ s^{-1}$,then the speed of the ball after $2 \ ms$ is: (in $m \ s^{-1}$)

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