A 2 kg ball moving at 24 ms^(-1) undergoes head on collision with a 4 kg ball moving in the opposite direction at 48 m s^(-1). If the coefficient of restitution is 2/3, their velocities in m s^(-1) after collision are

A 2 kg ball moving at 24 ms^(-1) undergoes head on collision with a 4

1.	A 2 kg ball moving at 24 ms^(-1) undergoes head on collision with a 4 kg ball moving in the opposite direction at 48 m s^(-1). If the coefficient of restitution is 2/3, their velocities in m s^(-1) after collision are
Answer» `- 56, - 8` `– 28, -4` `– 14, -2 ` `- 7,-1 ` SOLUTION :Here , `m_1 = 2kg , m_2 = 4kg = 2m_1` `u_1 = 24 MS^(-1) , u_2= -48 ms^(-1) = -2u_1` where `u_1` and `u_2` be the velocities of masses `m_1` and `m_2` before collision. According to the law of conservation of linear momentum , we GET `m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2v_2 , m_1 u_1 + 2m_1(-2u_1) = m_1 v_1 + 2m_1 v_2` `= -3 u_1 = v_1 + 2v_2 "".........(i)` Coefficient of restitution , `E = (v_2 - v_1)/(u_1 - u_2)` `v_2 - v_1 = e(u_1 - u_2)= e(u_1 - (-2u_1))` `v_2 - v_1 = 3u_1 e"" ....(ii)` Adding (i) and (ii), we get `3v_2 = 3u_1 e - 3u_1 " or " v_2 = u_1 (e - 1) "" .......(iii)` Substituting this value of `v_2` in (i), we get `v_1 = -u_1 (1 + 2e) "" ......(iv)` Substiting the given VALUES in (iii) and (iv), we get `v_2 = 24 (2/3 - 1) = -8 ms^(-1)` and `v_1 = -24(1 + 2 xx 2/3) = -56 ms^(-1)` .