A shell of mass 20 kg at rest explodes into two fragments whose masses are in the ratio 2 : 3. The smaller fragment moves with a velocity of 6 ms^-1. The kinetic energy of the larger fragment is

The correct answer is option 4) i.e. 96 J.

CONCEPT:

• Momentum: Momentum is the impact due to a moving object of mass, m, and velocity, v.

The momentum (p) of an object is expressed as:

p = mv

• Law of conservation of momentum: Momentum is conserved for any interaction between two objects occurring in an isolated system.
  • The conservation of momentum can be observed by analyzing the momentum of the total system or by analyzing the change in momentum.
  • This is done by equating the momentum before the interaction to the momentum after the interaction.
• Kinetic energy is the energy possessed by a moving object. Kinetic energy (KE) is expressed as:

KE = \(\frac{1}{2} mv^2\)

Where m is the mass of the object and v is the velocity of the object.

CALCULATION:

Given that:

Mass of the shell (m) = 20 kg

Let the mass of smaller fragment be m₁ and the mass of larger fragment be m₂.

Let the velocity of smaller fragment be v₁ (6 m/s) and velocity of larger fragment be v₂.

Ratio of masses of fragment =\(\frac{m_{1}}{m_{2}}\) = \(\frac{2}{3}\) \(⇒\) m₁ = \(\frac{2}{3}\)m₂   ----(1)

The mass of the shell = sum of the fragments

m = m₁ + m₂      ----(2)

Substituting (1) in (2),

20 = \(\frac{2}{3}\)m₂ + m₂

⇒ m₂ = 12 kg

Therefore, m₁ = m - m₂ = 20 - 12

⇒ m₁ = 8 kg

Using the principle of conservation of momentum:

Momentum_before = Momentum_after

m × v = (m₁ × v₁) + (m₂ × (-v₂))

( v₂ is taken -ve since the fragment is moving in opposite direction)

m × 0 = (8 × 6) + (12 × (-v₂))

( v = 0 m/s ∵ the shell is initially at rest)

0 = 48 - 12 v₂

⇒ v₂ = 4 m/s

Kinetic energy of the larger fragment = \(\frac{1}{2} m_{2}(v_{2})^2\) = \(\frac{1}{2} \times12\times4^2\) = 96 J