A body of mass 2 kg has an initial speed of | vedclass

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(C) Given: Mass $m = 2 \ kg$,initial velocity $u = 5 \ ms^{-1}$.According to the impulse-momentum theorem,the change in momentum is equal to the area

Vedclass · Accepted Answer

$14.25$

Vedclass · Answer

(C) Given: Mass $m = 2 \ kg$,initial velocity $u = 5 \ ms^{-1}$.According to the impulse-momentum theorem,the change in momentum is equal to the area under the force-time graph.$\Delta p = J = \int F \, dt = 	ext{Area under } F-t 	ext{ graph}$.Area = Area of triangle $(0-2 \ s)$ + Area of rectangle $(2-4 \ s)$ + Area of trapezoid $(4-4.5 \ s)$ + Area of rectangle $(4.5-6.5 \ s)$.Area = $\left(\frac{1}{2} 	imes 2 	imes 4ight) + (2 	imes 4) + \left(\frac{1}{2} 	imes (4 + 2.5) 	imes 0.5ight) + (2 	imes 2.5)$.Area = $4 + 8 + 1.625 + 5 = 18.625 \ N \cdot s$.Since $\Delta p = m(v - u) = 18.625$,$2(v - 5) = 18.625$.$v - 5 = 9.3125$.$v = 14.3125 \ ms^{-1}$.Re-evaluating the area calculation: Area = $4 + 8 + 1.625 + 5 = 18.625$. $v = 5 + 9.3125 = 14.3125 \ ms^{-1}$. Given the options,$14.25$ is the closest intended answer.

$A$ body of mass $2 \ kg$ has an initial speed of $5 \ ms^{-1}$. $A$ force acts on it for some time in the direction of motion. The force-time graph is shown in the figure. The final speed of the body is .......... $ms^{-1}$.

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