Two blocks A and B of mass mA and mB are connected together by means of a spring and are resting on a horizontal frictionless table. The blocks are then pulled apart so as to stretch the spring and then released. Show that the ratio of their kinetic energies at any instant is in the inverse ratio of their masses.
The Correct Answer and Explanation is:
To solve this problem, we need to use the principles of conservation of momentum and energy. The two blocks, A and B, of masses mAm_AmA and mBm_BmB, are connected by a spring on a frictionless surface. Initially, the blocks are pulled apart, and the spring is stretched. Upon release, the blocks move, and the spring exerts a force that accelerates both blocks.
Step 1: Conservation of Momentum
Since there are no external forces acting on the system, the total momentum of the system must remain constant. Initially, the blocks are at rest, so the initial momentum is zero. At any instant, let the velocities of blocks A and B be vAv_AvA and vBv_BvB, respectively. The conservation of momentum gives the equation:mAvA+mBvB=0m_A v_A + m_B v_B = 0mAvA+mBvB=0
This implies:mAvA=−mBvBm_A v_A = -m_B v_BmAvA=−mBvB
Thus, the velocities of the blocks are related by:vA=−mBmAvBv_A = -\frac{m_B}{m_A} v_BvA=−mAmBvB
Step 2: Energy Consideration
The total mechanical energy of the system is conserved. The potential energy stored in the spring at any instant is converted into the kinetic energy of the blocks. The total kinetic energy of the system is:K=12mAvA2+12mBvB2K = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2K=21mAvA2+21mBvB2
Using the relation vA=−mBmAvBv_A = -\frac{m_B}{m_A} v_BvA=−mAmBvB, we substitute vAv_AvA into the kinetic energy equation:K=12mA(mBmAvB)2+12mBvB2K = \frac{1}{2} m_A \left( \frac{m_B}{m_A} v_B \right)^2 + \frac{1}{2} m_B v_B^2K=21mA(mAmBvB)2+21mBvB2
Simplifying:K=12mA⋅mB2mA2vB2+12mBvB2K = \frac{1}{2} m_A \cdot \frac{m_B^2}{m_A^2} v_B^2 + \frac{1}{2} m_B v_B^2K=21mA⋅mA2mB2vB2+21mBvB2K=12(mB2mA+mB)vB2K = \frac{1}{2} \left( \frac{m_B^2}{m_A} + m_B \right) v_B^2K=21(mAmB2+mB)vB2
Step 3: Kinetic Energy Ratio
The kinetic energy of each block is given by:KA=12mAvA2andKB=12mBvB2K_A = \frac{1}{2} m_A v_A^2 \quad \text{and} \quad K_B = \frac{1}{2} m_B v_B^2KA=21mAvA2andKB=21mBvB2
Now, using vA=−mBmAvBv_A = -\frac{m_B}{m_A} v_BvA=−mAmBvB, we can find KAK_AKA:KA=12mA(mBmAvB)2=12mB2mAvB2K_A = \frac{1}{2} m_A \left( \frac{m_B}{m_A} v_B \right)^2 = \frac{1}{2} \frac{m_B^2}{m_A} v_B^2KA=21mA(mAmBvB)2=21mAmB2vB2
Thus, the ratio of the kinetic energies is:KAKB=12mB2mAvB212mBvB2=mBmA\frac{K_A}{K_B} = \frac{\frac{1}{2} \frac{m_B^2}{m_A} v_B^2}{\frac{1}{2} m_B v_B^2} = \frac{m_B}{m_A}KBKA=21mBvB221mAmB2vB2=mAmB
Therefore, the ratio of the kinetic energies of the two blocks is inversely proportional to the ratio of their masses, as required:KAKB=mBmA\frac{K_A}{K_B} = \frac{m_B}{m_A}KBKA=mAmB
Explanation
The key to solving this problem lies in understanding that the system is isolated, so both momentum and energy are conserved. By applying the conservation of momentum, we relate the velocities of the two blocks, and by using energy conservation, we establish the relationship between their kinetic energies. The result shows that the kinetic energy of each block is inversely proportional to its mass, meaning the block with a smaller mass will have a larger kinetic energy.
