Question

A projectile is fired at speed of $50\sqrt{2}\text{m/s}$ at an angle of ${45}^{\circ }$ from the horizontal. At $t=5\text{s}$, the projectile breaks into two parts of mass ratio $1:4$ and the lighter piece comes to rest immediately. Find the distance from the launching point to the point where the heavier piece lands.

• A. $250\text{m}$
• B. $750\text{m}$
• C. $500\text{m}$
• D. $562.5\text{m}$

Answer 1

The correct option is D $562.5\text{m}$


Horizontal component of velocity, $u_x=u\cos \theta =u_x=50\sqrt{2}\cos {45}^{\circ }=50\text{m/s}$

Lighter piece comes to rest after $5\text{s}$.
Therefore, Horizontal distance travelled by lighter piece
$=v_{x}t=50\times 5=250\text{m}$

Taking the point of projection as the reference point.
​​$x_{COM}=$ Range of projectile =$R$
$R=\frac{2u^2\sin \theta \cos \theta }{g}=\frac{2\times (50\sqrt{2}{)}^2\times 1/\sqrt{2}\times 1/\sqrt{2}}{10}=500\text{m}$

$\therefore x_{COM}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}=\frac{m\times 250+4m\times x_2}{4m+m}=\frac{250+4\times x_2}{5}$
$\Rightarrow 500=\frac{250+4\times x_2}{5}\Rightarrow 4\times x_2+250=5\times 500=2500$
$\Rightarrow x_2=\frac{2500-250}{4}=\frac{2250}{4}=562.5\text{m}$