Question

A mortar fires a shell of mass m which explodes into two pieces, m by 5, 4, m by 5, at the top of trajectory. The smaller mass falls very close to the mortar. In the same time, the bigger piece lands a distance d from the mortar. The shell would have fallen at a distance r from the mortar if there was no explosion. The value of d is.

Answer 1

5r/4

Answer 2

The problem can be solved by applying the principle of conservation of the motion of the center of mass. In the absence of external forces, the center of mass of a system continues to follow its original trajectory. The explosion is an internal force and does not affect the trajectory of the center of mass. Gravity is an external force, but it acts uniformly on all parts, meaning the center of mass will follow the same parabolic path as the original shell.

Let's define the coordinate system:

• The mortar is at the origin (x=0).
• The original shell, if it hadn't exploded, would have landed at a distance 'r' from the mortar. This means the final horizontal position of the center of mass of the system (the two pieces) will be 'r'.

Given:

• Total mass of the shell = m
• Mass of the smaller piece ($m_1$) = m/5
• Mass of the bigger piece ($m_2$) = 4m/5
• The smaller mass falls "very close to the mortar", so its landing position ($x_1$) = 0.
• The bigger piece lands at a distance 'd' from the mortar, so its landing position ($x_2$) = d.
• The final horizontal position of the center of mass ($X_{CM}$) = r.

The formula for the center of mass of a two-particle system is: $X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

Substitute the given values into the equation: $r = \frac{(\frac{m}{5}) \times 0 + (\frac{4m}{5}) \times d}{\frac{m}{5} + \frac{4m}{5}}$

Simplify the denominator: $\frac{m}{5} + \frac{4m}{5} = \frac{5m}{5} = m$

Simplify the numerator: $(\frac{m}{5}) \times 0 + (\frac{4m}{5}) \times d = 0 + \frac{4md}{5} = \frac{4md}{5}$

Now, substitute these simplified terms back into the equation: $r = \frac{\frac{4md}{5}}{m}$

Cancel 'm' from the numerator and denominator: $r = \frac{4d}{5}$

Finally, solve for 'd': $d = \frac{5r}{4}$

The value of d is $\frac{5r}{4}$.