Question

a bomb of mass 21 kgs kept at rest explodes in two parts in ratio of masses 2:5 , the lighter pert travels with a speed of 50 m/s then heavier mass travels with a speed (in m/s) of

Answer 1

Given : a bomb of mass 21 kgs kept at rest explodes in two parts in ratio of masses 2:5 ,

the lighter part travels with a speed of 50 m/s  

To Find : heavier mass travels with a speed (in m/s)

Solution:

bomb of mass 21 kgs

ratio of masses 2:5 ,

lighter part = (2/7)21 = 6 kg

Heavier part = (5/7)21  = 15 kg

Initial Momentum =   0   as it is  at rest

Final momentum  = 6 * 50 + 15v

Using law of conservation of momentum

6 * 50 + 15v = 0

=> 15v = - 300

=> v = - 20

Hence  heavier mass travels with a speed 20  m/s in opposite direction to lighter weight

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Answer 2

Given:

A bomb of mass 21 kgs kept at rest explodes in two parts in ratio of masses 2:5 . The lighter part travels with a speed of 50 m/s.

To find:

Speed of heavier mass ?

Calculation:

• Mass of lighter part is 21 × 2/(2+5) = 6 kg.

• Mass of heavier part = 21 × 5/(2+5) = 15 kg.

During fragmentation of the bomb, there was no external force. Hence the total momentum of system is conserved.

Applying CONSERVATION OF LINEAR MOMENTUM:

$Mu_{1} = m_{1}v_{1} + m_{2}v_{2}$

$\implies M(0)= 6(50)+ 15(v_{2})$

$\implies 0= 300+ 15(v_{2})$

$\implies 15(v_{2}) = - 300$

$\implies v_{2} = - 20 \: m {s}^{ - 1}$

So, speed of heavier part will be 20 m/s opposite to the direction of lighter part.