A projectile of mass 3m explodes at the highe | vedclass

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(C) Let the total mass be $M = 3m$. The projectile is launched from the origin $(0,0)$. The range $R = 100\,m$. The highest point is at $x = R/2 = 50\

Vedclass · Accepted Answer

$250$

Vedclass · Answer

(C) Let the total mass be $M = 3m$. The projectile is launched from the origin $(0,0)$. The range $R = 100\,m$. The highest point is at $x = R/2 = 50\,m$.After the explosion at the highest point,the center of mass continues to follow the original parabolic path and lands at $x = R = 100\,m$.Let the three parts be $m_1 = m_2 = m_3 = m$. Part $1$ retraces its path,so it lands at $x_1 = 0$.Part $2$ comes to rest at the highest point,so it falls vertically and lands at $x_2 = 50\,m$.Part $3$ lands at some position $x_3 = x$.The center of mass position $X_{cm}$ is given by the formula: $X_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{m_1 + m_2 + m_3}$.Substituting the known values: $100 = \frac{m(0) + m(50) + m(x)}{3m}$.$100 = \frac{50 + x}{3}$.$300 = 50 + x$.$x = 250\,m$.

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