A projectile of mass 3m is moving at 40, m/s | vedclass

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(A) At the highest point,the initial momentum of the projectile is $P_i = (3m)(40) = 120m$ in the horizontal direction.Let the velocity of the part wi

Vedclass · Accepted Answer

$130$

Vedclass · Answer

(A) At the highest point,the initial momentum of the projectile is $P_i = (3m)(40) = 120m$ in the horizontal direction.Let the velocity of the part with mass $m$ be $\vec{v} = v_x \hat{i} + v_y \hat{j}$.The part with mass $2m$ moves vertically up with velocity $25\, m/s$,so its momentum is $\vec{P}_{2m} = (2m)(25\hat{j}) = 50m\hat{j}$.By the law of conservation of linear momentum,$\vec{P}_i = \vec{P}_{2m} + \vec{P}_m$.$120m\hat{i} = 50m\hat{j} + m(v_x \hat{i} + v_y \hat{j})$.Equating the components:Horizontal: $120m = mv_x \implies v_x = 120\, m/s$.Vertical: $0 = 50m + mv_y \implies v_y = -50\, m/s$.The speed of the part with mass $m$ is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{120^2 + (-50)^2} = \sqrt{14400 + 2500} = \sqrt{16900} = 130\, m/s$.

$A$ projectile of mass $3m$ is moving at $40\, m/s$ at its highest point,where it breaks into two parts $m$ and $2m$. Mass $2m$ moves vertically up at $25\, m/s$. The other part will move at a speed of .................... $m/s$.

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