Question

At an amusemlec 27 collisions and powerent park, a 96.0 kg car moving with a speed v bounces elastically off a 135-kg bumper car at rest. If the final speed of the 135-kg car is 1.03 m/s, what is the initial speed of the 96.0-kg car?

Answer 1

By conservation of momentum :

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\96v + 0 = 96v_1 + 135\times 1.03\\\\96v = 96v_1 +139.05 \ \ \ \ \ \ .....1$

By conservation of energy :

$\dfrac{m_1u_1^2}{2}+\dfrac{m_1u_2^2}{2} =\dfrac{m_2v_1^2}{2} +\dfrac{m_2v_2^2}{2}\\\\96v^2 + 0 = 96v_1^2 + 135(1.03)^2\\\\96v^2 = 96v_1^2 +143.22 \ \ \ \ \ ....2)$

Solving equation 1 and 2, we get :

v = 1.23 m/s

Therefore, the initial speed of 96 kg car is 1.23 m/s .