A block of mass 4 kg is sliding on a frictionless horizontal surface with a kinetic energy of 18 J . It strikes a rigid wall perpendicularly and rebounds in the opposite direction with a kinetic energy of 8 J . The magnitude of the impulse imparted to the block by the wall is :

Step-by-step solution

Initial kinetic energy K_i = 18 J Magnitude of initial momentum p_i = 2mK_i = 2 4 18 = 12 kg m s ⁻¹ Final kinetic energy K_f = 8 J Magnitude of final momentum p_f = 2mK_f = 2 4 8 = 8 kg m s ⁻¹ Since the block rebounds in the opposite direction, the initial and final momenta have opposite signs. Magnitude of impulse J = | p | = |p_f - (-p_i)| = p_f + p_i J = 8 + 12 = 20 kg m s ⁻¹ Answer: 20 kg m s ⁻¹