An 8.00-kg ball, hanging from the ceiling by a light wire 135 cm long, is struck in an elastic collision by a 2.00-kg ball moving horizontally at 5.00 m/s just before the collision. Find the tension in the wire just after the collision.

Solution 89P Step 1 of 6: In the given problem, a ball of mass m = 2kg moving with initial speed u = 5 m/s 1 1 undergoes a collision with a ball of mass m =8 kg hanging with light wire of length 2 L=1.35 m, hence speed of hanging ball is u = 0 and after collision both balls move in 2 opposite direction with final speed of ball v 1nd hanging ball with v , 2s shown in the figure below we need to calculate the tension in wire after collision Step 2 of 6: Since it is elastic collision, both linear momentum and kinetic energy will be conserved. By conservation of momentum(P=mv), Momentum before collision = Momentum after collision m 1 1 m v 2 2 v 1 1 (as after collision balls move in opposite direction) Solving for v 1 m2 v1= m v 2 u …1….1 1 Step 3 of 6: 1 2 By conservation of kinetic energy(KE= mv ), 2 kinetic energy before collision = kinetic energy after collision 1m u = m v + m v2 1 2 2 1 1 2 1 1 2 2 2 m u = v +2 2v 2 1 1 m1 2 Solving for u 1 2 2 m2 2 u 1 v +1 m v2 1 Substituting equation 1 in above equation, m 2 m u = ( 2v u ) + 2v2 1 m 1 2 1 m 1 2 2 m 2 2 2 m2 m2 2 u 1 ( m v2) + (u 1 2 m v2 1+ m v2 1 1 1 m m m 0 = ( 2v ) 2 2v u + 2v2 m 1 2 m1 2 1 m1 2 m (1+ 2 )v =2u m1 2 1