Question

A 5kg shell kept at rest suddenly splits up into three part . If two parts of mass 2kg each are found flying due north and east with a velocity of 5m/s each , what is the velocity of the third part after explosion

Answer 1

Given:

A 5kg shell kept at rest suddenly splits up into three part . If two parts of mass 2kg each are found flying due north and east with a velocity of 5m/s each.

To find:

The velocity of the third part after explosion ?

Calculation:

In this type of questions , you need to apply the principle of CONSERVATION OF LINEAR MOMENTUM in both X and Y axis:

Along X axis: (all the X component velocity are to be used)

$\therefore \: P1 = P2$

$\implies \: mu = m1v1 + m2v2 + m3v3$

$\implies \: 5(0) = 2(5) +2(0)+ 1(v_{x})$

$\implies \: (v_{x}) = - 10 \: m {s}^{ - 1}$

Along Y axis: (all the Y component velocity are to be used)

$\therefore \: P3 = P4$

$\implies \: mu = m1v1 + m2v2 + m3v3$

$\implies \: 5(0) = 2(0) +2(5)+ 1(v_{y})$

$\implies \: (v_{y}) = - 10 \: m {s}^{ - 1}$

So, net velocity of 1 kg part :

$\therefore \: v = \sqrt{ {( v_{x}) }^{2} + {( v_{y}) }^{2} }$

$\implies \: v = \sqrt{ {(-10) }^{2} + {(-10) }^{2} }$

$\implies \: v = \sqrt{200}$

$\implies \: v = 10 \sqrt{2} \: m {s}^{ - 1}$

So, velocity of 3rd fragment is 10√2 m/s.