Question

2, A 20 gram particle traveling to the right with a speed of 40 cm/sec strikes a 40 gram particle traveling to the lef with a speed of 20 cm/sec. After the collision, the two particles move along the line of coincidence. If the collision is perfectly elastio (mechanical energy is conserved), what are the final velocities of the particles?

Answer 1

Since, elastic collision, Linear momentum conserved, Thus m_1v_1 + m_2v_2 = m_1 v_g1 + m_2 v_g2 where v_g1 & v_g2 are final velocity of mass m_1 & m_2 respectively or, 20 times 40 + 40(- 20) = 20 v_g1 + 40 v_g2 Rightarrow v_g1 + 2v_g2 = 0 Rightarrow v_g1 = -2 v_g2 Also kinetic energy is conserved 1/2 m_1 v^2_1 + 1/2 m_2 v^2_2 1/2 mv^2_g_1 + 1/2 mv^2_g_2 or, 1/2(20 times 40^2 + 40(- 20)^2) = 1/2(20v^2_g_1 + 40 v^2_g_2) or, 32000 + 16000 = 20(- 2 v^2_g_1)^2 + 40 v^2_g_2 Rightarrow v^2_g_2 = 48000/120 = 400 Therefore v_g_2 = squareroot 400 = 20 cm/s and v_g_1 = -2 v_g_2 = -2 times 20 = -40 cm/s velocity remains same just direction has reversed.