A particle of mass 0.02 kg moving with a speed of 100 m s⁻¹ strikes the rim of a uniform disc mounted on a smooth fixed axle through its centre. After the collision, the particle continues in the same direction with a speed of 90 m s⁻¹, while the disc starts rotating with an angular speed of 15.62 rad s⁻¹. The moment of inertia of the disc about its centre is 0.06 kg m².

Amount of energy loss (in J) in the collision is:

The correct answer is 11.68 J.

We use the law of conservation of energy to find the energy lost. The energy loss in any collision equals the difference between the total initial kinetic energy and the total final kinetic energy of the system.

Given Data:
Mass of particle: m = 0.02 kg
Initial speed of particle: u = 100 m s⁻¹
Final speed of particle: v = 90 m s⁻¹
Angular speed of disc after collision: ω = 15.62 rad s⁻¹
Moment of inertia of disc: I = 0.06 kg m²

Step 1: Calculate the Initial Kinetic Energy

Before the collision, only the particle is moving. The disc is at rest, so it contributes zero kinetic energy.

{\mathrm{KE}}_{\mathrm{initial}}=\frac{1}{2}m{u}^2=\frac{1}{2}\times 0.02\times {100}^2

{\mathrm{KE}}_{\mathrm{initial}}=\frac{1}{2}\times 0.02\times 10000=100\text{J}

Step 2: Calculate the Final Kinetic Energy of the Particle

After the collision, the particle continues in the same direction but with a reduced speed of 90 m s⁻¹.

{\mathrm{KE}}_{\mathrm{particle},\mathrm{final}}=\frac{1}{2}m{v}^2=\frac{1}{2}\times 0.02\times {90}^2

{\mathrm{KE}}_{\mathrm{particle},\mathrm{final}}=\frac{1}{2}\times 0.02\times 8100=81\text{J}

Step 3: Calculate the Final Rotational Kinetic Energy of the Disc

The collision sets the disc spinning. Its rotational kinetic energy is given by \frac{1}{2}I{\omega }^2.

{\mathrm{KE}}_{\mathrm{disc},\mathrm{final}}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\times 0.06\times {\mathrm{(15.62)}}^2

{\mathrm{KE}}_{\mathrm{disc},\mathrm{final}}=0.03\times 243.9844=7.32\text{J}

Step 4: Calculate the Total Final Kinetic Energy

{\mathrm{KE}}_{\mathrm{final}}={\mathrm{KE}}_{\mathrm{particle},\mathrm{final}}+{\mathrm{KE}}_{\mathrm{disc},\mathrm{final}}=81+7.32=88.32\text{J}

Step 5: Calculate the Energy Loss

\Delta E={\mathrm{KE}}_{\mathrm{initial}}-{\mathrm{KE}}_{\mathrm{final}}=100-88.32

\Delta E=11.68\text{J}

The energy is lost due to the inelastic nature of the collision — some of the particle's kinetic energy is permanently converted to heat, sound, and deformation at the point of impact. The amount of energy lost in the collision is 11.68 J.