Question

A projectile is fired at a speed of 100 m/s at an angle of ${37}^{\circ }$ above the horizontal. At the highest point, the projectile breaks into two parts of mass ratio 1 : 3, the lighter piece coming to rest. Find the distance from the launching point to the point where the heavier piece lands.

• A. 960 m
• B. 1000 m
• C. 1120 m
• D. 480 m

Answer 1

The correct option is C 1120 m

Answer is C.
Internal force do not effect the motion of the center of mass, the center of mass hits the ground at the position where the original projectile would have landed. The range of the original projectile is,
$x_{COM}=\frac{2u^{2}sin\theta cos\theta }{g}=\frac{2\times {10}^4\times \frac{3}{5}\times \frac{4}{5}}{10}m$ = $=960m$.
The center of mass will hit the ground at this position. As the smaller block comes to rest after breaking, it falls down vertlcally and hits the ground at half of the range, l.e., at x = 480 m. If the heavier block hits the ground at $x_2$, then
$x_{COM}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$.
$960=\frac{\left(m\right)\left(480\right)+\left(3m\right)\left(x_2\right)}{(m+3m)}$
$x_2=1120m$.
Hence, the distance from the launching point to the point where the heavier piece lands is 1120 m.