Question

A shell in flight explodes into n equal fragments k of the fragments reach the
ground earlier than the other fragments. The acceleration of their centre of mass subsequently will be:

• A. $g$
• B. $\left(nk\right)g$
• C. $\frac{(n-k)g}{k}$
• D. $\frac{(n-k)}{n}g$

Answer 1

The correct option is D $\frac{(n-k)}{n}g$

Given that,

A shall in flight explodes into n equal fragments, k of the fragments reach the ground earlier than the other fragments.

Now,

The acceleration of center of mass is

$a_{cm}=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}$

$a_{cm}=\frac{k\times 0+(n-k)\times g}{n}$

$a_{cm}=\frac{(n-k)g}{n}$

The acceleration of their center of mass will be $\frac{(n-k)g}{n}$.

Hence, D is the correct option.