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MCQEasyJEE 2024Elastic & Inelastic Collisions

JEE Physics 2024 Question with Solution

Consider a disc of mass 5 kg5 \, \text{kg}, radius 2 m2 \, \text{m}, rotating with angular velocity of 10 rad/s10 \, \text{rad/s} about an axis perpendicular to the plane of rotation. An identical disc is gently placed over the rotating disc along the same axis. The energy dissipated so that both discs continue to rotate together without slipping is:

  • A

    200 J200 \, \text{J}

  • B

    250 J250 \, \text{J}

  • C

    300 J300 \, \text{J}

  • D

    350 J350 \, \text{J}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A disc of mass 5 kg5 \, \text{kg} and radius 2 m2 \, \text{m} rotates with angular velocity 10 rad/s10 \, \text{rad/s}. An identical stationary disc is placed over it, and both rotate together finally.

Find: The energy dissipated during the process.

This is an inelastic rotational interaction. Since there is no external torque about the common axis, angular momentum is conserved, while kinetic energy decreases.

For one solid disc, the moment of inertia about its central axis is

I=12MR2I = \frac{1}{2}MR^2

So,

I=12(5)(2)2=10 kg⋅m2I = \frac{1}{2}(5)(2)^2 = 10 \, \text{kg} \cdot \text{m}^2

Hence, for the two identical discs,

I1=I2=10 kg⋅m2I_1 = I_2 = 10 \, \text{kg} \cdot \text{m}^2

Initially, only the first disc rotates. Therefore, initial angular momentum is

Linitial=I1ωi=10×10=100 kg⋅m2/sL_{\text{initial}} = I_1 \omega_i = 10 \times 10 = 100 \, \text{kg} \cdot \text{m}^2/\text{s}

Initial rotational kinetic energy is

Kinitial=12I1ωi2=12(10)(10)2=500 JK_{\text{initial}} = \frac{1}{2}I_1\omega_i^2 = \frac{1}{2}(10)(10)^2 = 500 \, \text{J}

After contact, both discs rotate together with common angular velocity ωf\omega_f. The total moment of inertia becomes

Ifinal=I1+I2=20 kg⋅m2I_{\text{final}} = I_1 + I_2 = 20 \, \text{kg} \cdot \text{m}^2

Using conservation of angular momentum,

Linitial=LfinalL_{\text{initial}} = L_{\text{final}} 100=20ωf100 = 20\omega_f ωf=5 rad/s\omega_f = 5 \, \text{rad/s}

Now the final kinetic energy is

Kfinal=12Ifinalωf2=12(20)(5)2=250 JK_{\text{final}} = \frac{1}{2}I_{\text{final}}\omega_f^2 = \frac{1}{2}(20)(5)^2 = 250 \, \text{J}

Hence, energy dissipated is

Edissipated=Kinitial−KfinalE_{\text{dissipated}} = K_{\text{initial}} - K_{\text{final}} Edissipated=500−250=250 JE_{\text{dissipated}} = 500 - 250 = 250 \, \text{J}

Therefore, the energy dissipated is 250 J250 \, \text{J}. The correct option is B.

Half-energy shortcut

Given: Two identical discs, one rotating and the other initially at rest.

Find: The energy dissipated after they rotate together.

For two identical discs, the final angular velocity becomes half of the initial value because angular momentum is conserved and the moment of inertia doubles.

Initially,

Ki=12Iω2K_i = \frac{1}{2}I\omega^2

After sticking together, total moment of inertia is 2I2I and common angular velocity is ω/2\omega/2. So,

Kf=12(2I)(ω2)2=14Iω2K_f = \frac{1}{2}(2I)\left(\frac{\omega}{2}\right)^2 = \frac{1}{4}I\omega^2

Thus,

Kf=12KiK_f = \frac{1}{2}K_i

So exactly half of the initial kinetic energy is dissipated.

Here,

Ki=12(10)(10)2=500 JK_i = \frac{1}{2}(10)(10)^2 = 500 \, \text{J}

Therefore,

Edissipated=500−250=250 JE_{\text{dissipated}} = 500 - 250 = 250 \, \text{J}

Therefore, the energy dissipated is 250 J250 \, \text{J}. The correct option is B.

Common mistakes

  • Using conservation of kinetic energy is incorrect because friction between the discs makes the process rotationally inelastic. Conserve angular momentum, then compute the loss in kinetic energy.

  • Taking the final moment of inertia as that of a single disc is wrong. After both discs rotate together, the total moment of inertia is the sum of both discs: I1+I2I_1 + I_2.

  • Assuming the final angular velocity remains 10 rad/s10 \, \text{rad/s} is incorrect. Since angular momentum is shared by two identical discs, the common angular velocity decreases to 5 rad/s5 \, \text{rad/s}.

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