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.
