A block of clay, mass 5 kg, slides down a frictionless 32o incline that is 4.0 m

Question

A block of clay, mass 5 kg, slides down a frictionless 32o incline that is 4.0 m high and hits a 2 kg wooden block situated at the bottom of the incline. The collision is "perfectly" inelastic. The horizontal surface is frictionless for 1 meter, then it is rough for the remainder of the movement with a coefficient of kinetic friction (µk = 0.37). The system comes to rest after moving some distance, d, across the rough horizontal surface. (Assume the friction is the same for both clay and wood!)

Determine the distance, d, the clay and block slide from the bottom of the ramp. [Hint: This will require splitting up the problem into parts and using several concepts from current as well as past material]

Explanation / Answer

m*g*h = 0.5m*u^2

u = sqrt(2*g*h) = 8.54 m/s

during collision

m*u*cos32 = (M+m)*V

V = 5.17 m/s

on rough surface

f = uk*N = uk*(M+m)*g

0.5*(M+m)*V^2 - uk*(M+m)*g*d = 0

d = V^2/2*uk*g = 3.69 m

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