Question

A block with a mass m1=2.3kg is sliding along a frictionless surface with a velocity of 7.3m/s. It collides inelastically with mass m2=1.7kg and the two blocks stick together. They then slide down a frictionless incline with a Height 95cm. How fast are they going when they reach the bottom of the incline?

Part B. If the coefficient of kinetic friction, uk is 0.15 along the surface at the bottom of the ramp. What distance will the blocks side before coming to rest?

Answer 1

let

m1 = 2.3 kg, u1 = 7.3 m/s
m2 = 1.7 kg, u2 = 0
let vo is the combined speed after the collision and v is the combined speed at the bottom of the hill.
A)
Apply conservation of momentum

m1*u1 + m2*u2 = (m1 + m2)*vo

vo = (m1*u1 + m2*u2)/(m1 + m2)

= (2.3*7.3 + 0)/(2.3 + 1.7)

= 4.20 m/s

now apply conservation of energy, (1/2)*(m1+m2)*v^2 = (1/2)*(m1 + m2)*vo^2 = (m1 + m2)*g*h

v^2 = vo^2 + 2*g*h

v = sqrt(vo^2 +2*g*h)

= sqrt(4.2^2 + 2*9.8*0.95)

= 6.02 m/s <<<<<<<<<<<<<----------------Answer

b) acceleration of the blocks on rough surface,

a = -g*mue_k

= -9.8*0.15

= -1.47 m/s^2

distance tarvelled by the blocks before stopping,

d = (vf^2 -vi^2)/(2*a)

= (0^2 - 6.02^2)/(2*(-1.47))

= 12.3 m <<<<<<<<<<<<<----------------Answer