A cannon ball is fired with a velocity 200, m | vedclass

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(B) Let the mass of the cannon ball be $3m$. At the highest point,the vertical component of velocity is $0$,and the horizontal component is $v_x = u \

Vedclass · Accepted Answer

$300\, m/s$ in the horizontal direction

Vedclass · Answer

(B) Let the mass of the cannon ball be $3m$. At the highest point,the vertical component of velocity is $0$,and the horizontal component is $v_x = u \cos 	heta = 200 \cos 60^{\circ} = 200 	imes 0.5 = 100\, m/s$.By the law of conservation of linear momentum,the initial momentum at the highest point is $\vec{P}_i = (3m)(100\, \hat{i}) = 300m\, \hat{i}$.Let the velocities of the three fragments be $\vec{v}_1 = 100\, \hat{j}$,$\vec{v}_2 = -100\, \hat{j}$,and $\vec{v}_3$ respectively.According to the conservation of momentum: $\vec{P}_i = m\vec{v}_1 + m\vec{v}_2 + m\vec{v}_3$.$300m\, \hat{i} = m(100\, \hat{j}) + m(-100\, \hat{j}) + m\vec{v}_3$.$300m\, \hat{i} = 0 + m\vec{v}_3$.Therefore,$\vec{v}_3 = 300\, \hat{i}\, m/s$.The third fragment moves with a velocity of $300\, m/s$ in the horizontal direction.

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