Question

Question 8. (15 Marks) Two gliders move toward each other on a frictionless track (Figure O8a). Glider A has a mass of 0.50 kg, and glider B has a mass of 0.30 kg: both gliders move with an initial speed of 2.0 m/s. (a) Assume that after the collision, glider B moves away with a final velocity of +2.0 m/s (Figure Q8c). What is the final velocity of A? (b) Assume that after the collision, the two gliders stick together and move as one object (Figure Q8d). What is the final velocity of the joined gliders? (c) Find the ratio of the final to initial kinetic energies for part (b) VALE - 2.0 m/s UBL = -2.0 m/s - A MA -0.50 kg m - 0.30 kg (a) Before collision A (b) Collision UAGE 0B6X = 2.0 m/s (e) After collision (d) After collision Figure Q8

Answer 1

(A) Let initial velocity of the first sphere is Ver = 2.0m/s The initial velocity of the second sphere is Vai = -2.0 m/s The initial velocity of the second sphere is Vr = 2.0 m/s Let , and v, are the final velocities of masses m, and m respectively The masses are: m = 0.5 kg mg=0.3 kg As there is no extemal force acting on the system, linear momentum conserved. That is total initial momentum is equal to total final momentum. m Vari+mz Vaxi = m_V44 +mzVey m. (Vasi – 2.48)= mg (Vgy - Vai) ----(1) From conservation of energy m_(visi – v 2) = mg (wy - Vži) m. (Vari - V-7)(Vari + V27) = mg (vey - Vai)(Vxy + vži) ----(2) Divide equation (2) by equation (1). V&xi + V dr = Vzxy + Vert Vanvi = Voxy + Vexi - Vai----(3) vai = axi -Y/+Vi ----(4) Substitute equation 4 in equation 1 m. Vasi-V48) = mg (784 - Veri m. (vas – V24) = m;(( V.si – Vai +V44)-ver) _(0.5-0.3) 2+( 2(0.3) (-2) (0.5+0.3)(2)+(05703}(-2) =-1.0 m/s Substitute equation (3) in equation (1)

m. (V- Vx) = m (Vor - Vai) m_(Veri -(vey +v=V+)) = mz (voy – Vəti) V BXE Vem+mg ** (ma+m3) ( 2(0.5) (2)+(0.3-0.5) (-2) 0.5+0. 34*05+0.3(-2) = 3.0 m/s (B) The total initial momentum is equal to total final momentum in the case of inelastic collision. m vai + mg Vi = (m. +mg), m Veri+ mg Vi m. +mg (0.5)(2)+(0-3)(-2) 0.5+0.3 = 0.5 m/s V = (C) The ratio is 3(m +mg )v; (m+ m2) v m Vări + Mg Vedi (0.5+0.3)(0.5) (0.5)(2) +(0.3)(-2) = 0.0625