Question

a small sphere of mass m=2Kg moving with a velocity u=4i-7jm/s collides with a smooth wall and returns with a velocity v=-i+3jm/s. the magnitude of the impulse recived by the ball is.

Answer 1

Answer:

22.3 kg m/s

Explanation:

Impulse= Force* timeperiod=mass * (change in velocity)

Impluse = mass*(v-u)

             =2kg*(-i+3j -(4i-7j)) m/s

             =2kg*(-5i+10j)m/s

            =2kg*squareroot(125)m/s

           =2*(11.18)kgm/s

          =22.3kg m/s

Answer 2

here, we have to use formula,

impulse = change in momentum

impulse = Pf - Pi

impulse = m(Vf - Vi)

here, m = 2 kg

Vf = (-i + 3j ) m/s

Vi = (4i - 7j) m/s

now,

impulse = 2 × ( -i + 3j - 4i + 7j)

= 2 × ( -5i + 10j)

=(-10i + 20j ) kg.m/s

hence, magnitude of impulse = √(10²+20²)

= 10√5 N.s