Question

A 8.3-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?

Answer 1

Answer:

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.