A projectile of mass M is fired so that the horizontal range is 4 km. At the highest point the projectile explodes in two parts of masses M/4 and 3M/4 respectively and the heavier part starts falling down vertically with zero initial speed. The horizontal range (distance from point of firing) of the lighter part is :

Question

A projectile of mass M is fired so that the horizontal range is 4 km. At the highest point the projectile explodes in two parts of masses M/4 and 3M/4 respectively and the heavier part starts falling down vertically with zero initial speed. The horizontal range (distance from point of firing) of the lighter part is : (A) 16 km (B) 1 km (C) 10 km (D) 2 km

Tardigrade Admin · Accepted Answer

X COM =( m 1 x 1+ m 2 x 2/ m 1+ m 2) R =(( M /4) x +(3 M /4) × ( R /2)/ M ) ⇒ x =10 Km

16 km

1 km

10 km

2 km

XCOM​=m1​+m2​m1​x1​+m2​x2​​ R=M4M​x+43M​×2R​​ ⇒x=10Km

$X_{COM} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}}$
$R = \frac{\frac{M}{4} x + \frac{3 M}{4} \times \frac{R}{2}}{M}$
$\Rightarrow x = 10 K m$