Question

Sample Problem 2: A stone of mass 100 kg breaks into two parts such that momentums of the parts are equal and opposite. If the first part of mass 60 kg moves towards north with velocity 20 ms. What is the velocity of the second part?​

Answer 1

Solution :-

As per the given question ,

• Initially the stone of mass 100 kg is at rest

Now ,due to some internal forces the stone breaks into two parts  

• m 1 = 60 kg

• m 2 = 40 kg

The stone of m 1 moves towards the north with velocity = 20 m/s

As no external force is acting on the stone we can say that the total momentum of the system is conserved

By applying law of conservation of energy ,

⟹ Pi = Pf

⟹ 0 = m 1 x v 1 + m 2 x v 2  (∵ Pi = 0 )

⟹ m 2 x v 2 = - m 1x v 1

⟹ 40 x v 2 = - 60 x 20

⟹ 40 x v 2 = - 1200

⟹ v 2 = - 1200 / 40

⟹ v 2 = - 30 m /s  

∴The velocity of the second part is 30 m/s ( towards the south )

Answer 2

$\large\sf{Solution}$

Mass of the Second part = 100 - 60 = 40kg

Momentum of the first part p1 = m v

= 60 kg × 20 ms ‐¹ = 1200 kg m s - ¹

Momentum of the second part p2 =m v

= 40 × v

But it is given that momentum of the two parts are equal to and opposite Hence ,

40 × v = 1200

v= $\frac{1200}{1000}$

= 30 m s - ¹ towards south