A8.3-g wad of sticky clay is hurled horizontally at a 82-g wooden block initially at rest on a horizontal surface. the clay sticks to the block. after impact, the block slides 7.50 m before coming to rest. if the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact?

The speed of the clay before the impact was 106.35 m/s.

Explanation:

the only force doing work on the system is the frictional force, f, the work done by f is given by:

Wf = ΔK = Kf - Ki

The clay and the block will come to rest after sliding Δx = 7.50 m, if their intial speed is v and the combined mass is m and μ is the coefficient of friction and g is gravity,then:

         f×Δx = Ki

m×g×Δx×μ = 1/2×m×v^2

           v^2 = 2×g×Δx×μ

                  =  2×(9.8)×(7.50)×(0.650)

                  = 95.55

               v = 9.78 m/s

This is the veloty of clay and block after the clay hit the block.

if the velocity the clay and block attains after the impact is v and the initial speed of the clay is v1 and the mass is m and the speed of the block initially is V = 0 m/s and the mass is M, then according to the conservation of linear momentum:

m×v1 +M×V = v(m + M)

         m×v1 = v(m + M)

              v1 = v(m + M)/m

              v1 =  (9.78)(8.3×10^-3 + 82×10^-3)/(8.3×10^-3)

              v1 =  106.35 m/s

Therefore, the speed of the clay before the impact was 106.35 m/s.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west