Answer to Question #117497 in Mechanics | Relativity for Neo

Question #117497
Two blocks are free to slide along the frictionless wooden track ABC. The block of mass m1=5.0 kg is released from A which is 5.0 m above the ground. Protruding from its front end is the north pole of a strong magnet, attracting the south pole of an identical magnet embedded in the back end of the block of mass m2 =10.0 kg, initially at rest on the ground. The two blocks touches and start to move at B. Calculate the maximum height at C to which m2m1 rises after the inelastic collision.
Expert's answer

Explanations & Calculations

  • From A to B mechanical energy of the block 1 is conserved. By that the velocity of the moving block just before the collision is found. (Assume the magnetic forces are negligible & no work is done by it)

"\\qquad\\qquad\n\\begin{aligned}\n\\small E_p +E_k &= \\small E_p+E_k \\\\\n\\small(m_1gh +0) &=\\small(0+\\frac{1}{2}m_1u^2)\\\\\n\\small u &= \\small\\sqrt{2gh}\\\\\n&= \\small \\sqrt{2\\times9.8ms^{-2}\\times5m}\\\\\n&= \\small 7\\sqrt{2}ms^{-1}\n\\end{aligned}"

  • For the collision at B, the linear momentum of the entire system is conserved as no external forces are present & kinetic energy is not conserved due to the collision is completely inelastic.

"\\qquad\\qquad\n\\begin{aligned}\n\\small m_1u +m_20 &= \\small (m_1+m_2)v\\\\\n\\small v &= \\small \\frac{m_1u}{m_1+m_2}\\\\\n&= \\small \\frac{5kg\\times7\\sqrt{2}ms^{-1}}{(5kg+10kg)}\\\\\n&= \\small \\frac{7\\sqrt{2}}{3}ms^{-1}\n\\end{aligned}"

  • Now the 2 block segment has some kinetic energy which is employed in taking the height ahead. That height could be found by applying conservation of mechanical energy from B to C.

"\\qquad\\qquad\n\\begin{aligned}\n\\small E_{k(m_1+m_2)} &= \\small \\frac{1}{2}(m_1+m_2)v^2\\\\\n\\end{aligned}"


"\\qquad\\qquad\n\\begin{aligned}\n\\small E_k+E_p &= \\small E_k+E_p\\\\\n\\small \\frac{1}{2}(m_1+m_2)v^2+0&=\\small 0+(m_1+m_2)gh\\\\\n\\small h &= \\small \\frac{v^2}{2g}\\\\\n\\small h &= \\small \\frac{(\\frac{7\\sqrt{2}}{3}ms^{-1})^2}{2\\times9.8ms^{-2}}\\\\\n&= \\small \\bold{0.56m}\n\\end{aligned}"

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