If a body of mass 2 ,kg moving with an initia | vedclass

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(B) Given: Mass $m = 2 \,kg$, initial velocity $u = 4 \,m \,s^{-1}$, force $F = 3 \,N$, time $t = 2 \,s$.Since the force is applied normal to the init

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$5 \,m \,s^{-1}$

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(B) Given: Mass $m = 2 \,kg$, initial velocity $u = 4 \,m \,s^{-1}$, force $F = 3 \,N$, time $t = 2 \,s$.Since the force is applied normal to the initial velocity, the initial velocity is along the $x$-axis ($u_x = 4 \,m \,s^{-1}$, $u_y = 0$).The acceleration produced by the force is $a = F/m = 3/2 = 1.5 \,m \,s^{-2}$.This acceleration acts in the $y$-direction, so $a_y = 1.5 \,m \,s^{-2}$ and $a_x = 0$.The final velocity components after $t = 2 \,s$ are:$v_x = u_x + a_x t = 4 + 0 = 4 \,m \,s^{-1}$.$v_y = u_y + a_y t = 0 + (1.5)(2) = 3 \,m \,s^{-1}$.The resultant velocity $v$ is given by $v = \sqrt{v_x^2 + v_y^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,m \,s^{-1}$.

If a body of mass $2 \,kg$ moving with an initial velocity of $4 \,m \,s^{-1}$ is subjected to a force of $3 \,N$ for a time of $2 \,s$ normal to the direction of its initial velocity, then the resultant velocity of the body is

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