A particle of mass 1, kg is acted upon by a f | vedclass

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(B) The impulse $J$ delivered to the particle is equal to the area under the force-time $(F-t)$ graph.Impulse $J = \Delta p = m(v_f - v_i)$.From the g

Vedclass · Accepted Answer

$110$

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(B) The impulse $J$ delivered to the particle is equal to the area under the force-time $(F-t)$ graph.Impulse $J = \Delta p = m(v_f - v_i)$.From the graph, the area from $t = 0$ to $t = 10\, s$ is a triangle with base $10\, s$ and height $20\, N$.Area $A_1 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 10 	imes 20 = 100\, Ns$.This area is positive, so the velocity increases during this interval.At $t = 10\, s$, the velocity $v_{10}$ is given by $m(v_{10} - v_i) = A_1$.$1 	imes (v_{10} - 10) = 100 \implies v_{10} = 110\, ms^{-1}$.From $t = 10\, s$ to $t = 20\, s$, the force is $-10\, N$ (a rectangle).Area $A_2 = 	ext{base} 	imes 	ext{height} = (20 - 10) 	imes (-10) = 10 	imes (-10) = -100\, Ns$.Since the force is negative, the velocity decreases after $t = 10\, s$.Therefore, the maximum velocity is attained at $t = 10\, s$, which is $110\, ms^{-1}$.

$A$ particle of mass $1\, kg$ is acted upon by a force $F$ which varies as shown in the figure. If the initial velocity of the particle is $10\, ms^{-1}$, the maximum velocity attained by the particle during the period is ........ $ms^{-1}$.

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