Question

A railway truck of 100kg mass traveling with a velocity of 7ms^-1collides with a second truck of 200kg mass and the two couple automatically and move off together. Calculate the velocity of the coupled trucks and the energy lost to the system; if the second truck has a speed of 5ms^-1(a) in the same direction, (b) in the opposite direction and (c) is initially stationary.

Answer 1

The conservation of momentum law is $\vec{p}_1 + \vec{p_2} = \vec{p}_{\text{tot}}$ or $m_1\vec{v}_1 + m_2\vec{v_2} = (m_1+m_2)\vec{v}_{\text{tot}}$

a) In case of the same direction

$m_1{v}_1 + m_2{v_2} = (m_1+m_2){v}_{\text{tot}}, \\ v_{\text{tot}} = \dfrac{m_1{v}_1 + m_2{v_2}}{m_1+m_2} = \dfrac{100\cdot7 + 200\cdot5}{100+200} = 5.67\,\mathrm{m/s}.$

The initial total kinetic energy is $0.5m_1{v_1}^2 + 0.5m_2{v_2}^2 = 4950\,\mathrm{J}.$

The final kinetic energy is $0.5(m_1+m_2)\cdot v_{\text{tot}}^2 = 4816.7\,\mathrm{J}.$

The energy lost is 133 J.

b) In case of the opposite directions

$m_1{v}_1 - m_2{v_2} = (m_1+m_2){v}_{\text{tot}}, \\ v_{\text{tot}} = \dfrac{m_1{v}_1 - m_2{v_2}}{m_1+m_2} = \dfrac{100\cdot7 - 200\cdot5}{100+200} = -1\,\mathrm{m/s}.$ So the final velocity is directed along the velocity of the second truck.

The final kinetic energy is $0.5(m_1+m_2)\cdot v_{\text{tot}}^2 = 150\,\mathrm{J}.$

The energy lost is 4800 J.

c) In case of initially stationary truck

$m_1{v}_1 = (m_1+m_2){v}_{\text{tot}}, \\ v_{\text{tot}} = \dfrac{m_1{v}_1 }{m_1+m_2} = \dfrac{100\cdot7 }{100+200} = 2.33\,\mathrm{m/s}.$

The final kinetic energy is $0.5(m_1+m_2)\cdot v_{\text{tot}}^2 = 816.6\,\mathrm{J}.$

The energy lost is 4133.3 J.