Question

If a shell fired from a cannon, explodes in mid-air then:

• A. Its total kinetic energy decreases
• B. Its total kinetic energy increases
• C. Its total kinetic momentum increases
• D. Its total kinetic momentum decreases

Answer 1

The correct option is B

Its total kinetic energy increases

Step 1: Given

Assume the shell splits into two in mid-air,

Initial mass$=m$

Initial velocity$=v$

Final mass$=\frac{m}{2}and\frac{m}{2}$

Final velocity $=v_{1}and{v}_2$

Step 2: Determine the velocity

As all the force is external, momentum is conserved.

$$\begin{gathered} mv=\frac{m}{2}(-v_1)+\frac{m}{2}{v}_2 \\ v=\frac{-v_1}{2}+\frac{v_2}{2} \\ 2v=v_2-v_1 \end{gathered}$$

Step 3: Determine the change in kinetic energy

$$\begin{gathered} ∆KE=K_f-K_i \\ =(\frac{1}{2}\frac{m}{2}{v_1}^2+\frac{1}{2}\frac{m}{2}{v_2}^2)-\left(\frac{1}{2}m{v}^2\right) \\ =\frac{m}{4}({v_1}^2+{v_2}^2)-\frac{1}{2}{(v_2-v_1)}^2=\frac{m}{4}(2v_1{v}_2+2{v_2}^2) \end{gathered}$$

$∆KE>0\Rightarrow$The kinetic energy has increased.

Hence, option (B) is the correct option.