Question

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.

Answer 1

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)

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

• 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.

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

• 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.

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

And,

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