A block of mass 2 kg is projected with an initial speed of 14 m/s on a rough horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.2 . After traveling a distance of 24 m , the block collides head-on with a massive stationary bumper. It rebounds and travels a distance of 9 m before coming to rest. What is the magnitude of the impulse imparted by the bumper to the block? (Take g

Step-by-step solution

The deceleration of the block due to kinetic friction is given by: a = - _k g = -0.2 10 = -2 m/s ^2 Let v₁ be the velocity of the block just before it collides with the bumper. Using the third equation of motion: v₁^2 = u^2 + 2a d₁ v₁^2 = (14)^2 + 2(-2)(24) v₁^2 = 196 - 96 = 100 v₁ = 10 m/s Let v₂ be the velocity of the block just after rebounding. Since it comes to rest after traveling d₂ = 9 m : 0 = v₂^2 + 2a d₂ v₂^2 = 2(2)(9) = 36 v₂ = 6 m/s The impulse J imparted by the bumper is equal to the change in momentum