Question

(b) A bomb weighing 50 kg explodes into three
parts in flight when its velocity is
20 î+22j +10k ms-1. Two fragments of the
bomb weighing 10 kg and 20 kg are found to
have velocities 100 i + 50j +20k and
30 î – 20j – 10Â ms-l respectively. Find the
velocity of the third fragment.​

Answer 1

The  velocity of the third fragment is $-30\hat{i}+50\hat{j}+25\hat{k}$

Explanation:

1. We know that during explosion total momentum is conserved.

  means

 initial momentum = final momentum

2. $M\vec{u}=M_{1}\vec{v_{1}}+M_{2}\vec{v_{2}}+M_{3}\vec{v_{3}}$    ...1)

3. $\vec{u}$= initial velocity in x- direction =$(20\hat{i}+22\hat{j}+10\hat{k})$ $(\frac{m}{s})$

$\vec{v_{1}}$ = final velocity of mass 10 kg =$(100\hat{i}+50\hat{j}+20\hat{k})$

  $\vec{v_{2}}$ = final velocity of mass 20 kg =$(30\hat{i}-20\hat{j}-10\hat{k})$

$\vec{v_{3}}$= final velocity of third mass 20 kg = $(v_{x}\hat{i}+v_{y}\hat{j}+v_{z}\hat{k})$

4. From equation 1)

   $50 (20\hat{i}+22\hat{j}+10\hat{k})= 10(100\hat{i}+50\hat{j}+20\hat{k})+20(30\hat{i}-20\hat{j}-10\hat{k})+20(v_{x}\hat{i}+v_{y}\hat{j}+v_{z}\hat{k})$

5. On comparing corresponding term

  First $\hat{i}$

   50×20= 10×100+20×30+20$v_{x}$

  So $v_{x}$= -30$(\frac{m}{s})$

  Second $\hat{j}$

 50×22= 10×50-20×20+20$v_{y}$

 So $v_{y}$= 50$(\frac{m}{s})$

Third $\hat{k}$

 50×10= 10×20-20×10+20$v_{z}$

 So $v_{z}$= 25$(\frac{m}{s})$

6. So velocity of third fragment = $-30\hat{i}+50\hat{j}+25\hat{k}$