Question

A ball of mass 0.2 kg is thrown against the wall . the ball strikes the wall normally with velocity of 30m/sand rebounds with velocity of 20m/s. calculate the impulse of the force exerted by the ball on the wall

Answer 1

m = 0.2 kg 
u = 30 m/s 
v =  20 m/s

impulse = change in the momentum 
             final momentum - initial momentum
             mv - mu
             m ( v - u )
        => 0.2 *  (-20 -30 )
             = -10 Ns     

Answer 2

given , u = 30 m/s , v = 20 m/s 
          m =  0.2 kg
          

impulse - change in momentum of the body 
             = mv - mu = m (v - u )
                            = 0.2 ( -20 -30 ) = -10 N