91Ó°ÊÓ

The initial velocity of the shell is$v_0=20m/s$

The inclination angle with the horizontal is $\mathrm{θ}=60°$

The speed of the one fragment after an explosion in the vertical direction is $v_{0y}=0m/s$

The law of conservation of momentum is stated as, if a system is isolated so that no net external force acts on it, the linear momentum of the system remains constant.We can use the law of conservation of the momentum and find the velocity of the fragment after the explosion. We can use the kinematic equations to find the distance covered.

Formula:

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

…(¾±)

$v_f=v_0+at$ …(¾±¾±)

$x-x_0=v_{0x}t$ …(¾±¾±¾±)

$y-y_0=v_{0y}t-\frac{1}{2}g{t}^2$ …(¾±±¹)

The horizontal distance between the second fragment of the land and the gun:

We can find the coordinates of the shell at the explosion point. Using equation (ii) and g as gravitational acceleration,

$$\begin{gathered} v_y=v_{0y}-gt \\ \left(0m/s\right)=v_{0y}-gt \\ t=\frac{v_{0y}}{g} \end{gathered}$$

The horizontal distance covered by the shell up to the highest point is

$\begin{array}{rcl}{x}_0& =& v_{0x}t\\ & =& v_{0x}\frac{v_{0y}}{g}\\ & & \end{array}$

Substituting the given values,

$\begin{array}{rcl}{x}_0& =& v_{0x}\frac{v_{0y}}{g}\\ & =& \frac{v_0\cos \mathrm{θ}×v_0\sin \mathrm{θ}}{g}\\ & =& \frac{\left(20\text{m/s}\right)\cos \left(60°\right)×\left(20\text{m/s}\right)\sin \left(60°\right)}{9.8{\text{m/s}}^{\text{2}}}\\ & =& 17.7\text{m}\\ & & \end{array}$

The vertical distance covered by the shell up to the highest point is

$y_0=v_{0y}t-\frac{1}{2}g{t}^2$

Substitute the value of t .

$\begin{array}{rcl}{y}_0& =& v_{0y}×\frac{v_{0y}}{g}-\frac{1}{2}g{\left(\frac{v_{0y}}{g}\right)}^2\\ & =& \frac{1}{2}\frac{{\left(v_{0y}\right)}^2}{g}\\ & =& \frac{1}{2}\frac{{\left(v_0\text{sin}\mathrm{θ}\right)}^2}{g}\\ & =& \frac{1}{2}\frac{{\left(20m/s×\sin \left(60°\right)\right)}^2}{9.8m/s^2}\\ & =& 15.3m\\ & & \end{array}$

The x and y coordinates of the shell at the explosion are $\left(x_0,y_0\right)=\left(17.7m,15.3m\right)$. The horizontal momentum is conserved. At the highest point, the velocity of the shell is before the explosion. After the explosion is the velocity of the second fragment which is moving along the positive x axis.

From the equation (i),

$\begin{array}{rcl}M{v}_{0x}& =& 0+\left(M/2\right)v_x\\ v_{0x}& =& 0+\left(1/2\right)v_x\\ v_x& =& 2v_{0x}\\ & =& 2×v_0\text{cos}\mathrm{θ}\\ & =& 2×20m/s×\text{cos}60\\ & =& 20m/s\\ & & \end{array}$

We can find the horizontal distance covered by the second fragment with projectile motion. One of the fragments falls directly in the vertical direction hence, its vertical initial velocity is zero. Hence, according to the equation (iv),

$\begin{array}{rcl}y-y_0& =& v_{0y}t-\frac{1}{2}g{t}^2\\ & =& -\frac{1}{2}g{t}^2\\ t^2& =& \frac{2\left(y_0-y\right)}{g}\\ t& =& \sqrt{\frac{2\left(y_0-y\right)}{g}}\\ & & \end{array}$

The horizontal distance covered by the second fragment can be calculated using equation (iii).

$\begin{array}{rcl}x-x_0& =& v_{x}t\\ x& =& x_0+v_x\sqrt{\frac{2\left(y_0-y\right)}{g}}\\ & =& x_0+v_x\sqrt{\frac{2\left(y_0-y\right)}{g}}\\ & =& 17.7m+\left(20m/s\right)\sqrt{\frac{2\left(15.3m-0\right)}{9.8m/s^2}}\\ & =& 53m\\ & & \end{array}$

Therefore, thehorizontal distance between the second fragment of land and the gun is $53m$.