  Question

# A box of mass 2 kg slides from a height of 8 m on the inclined plane and struck the ball. After collision, the ball sticks on the box. If mass of the ball is 200 g, then what will be the velocity of the box after collision? Assuming all surfaces are frictionless and initially the ball is at rest. (take g=10m/s2) 2.3 m/s4.5 m/s8.5 m/s11.5 m/s

Solution

## The correct option is D 11.5 m/sLet mass of the box = m1 mass of the ball  = m2 Given height h = 8 m Let velocity of box at point C, just before the collision = u1  Velocity of ball before collision u2=0 (At rest) Initially box has potential energy at point A and after reaching at the point C, just before the collision, potential energy will convert into kinetic energy. Applying energy conservation at point A and B. (P.E)at point A=(K.E)at point C. m1gh=12m1u21 u1=√2gh u1=√2×10×8 u1=√160 m/s u1=12.65 m/s. (Velocity of box in horizontal direction just before the collision) Let the velocity of box and ball after collision be v and applying the conservation of momentum before collision and after collision, m1u2+m2u2=(m1+m2)v ⇒2×12.65+(0.200)×0=(2+0.200)v ⇒v=2×12.652.2 ⇒v=11.5 m/s So, velocity of box after collision is 11.5 m/s.  Suggest corrections   