Question

Object 1 is held at rest at the top of a rough inclined plane of length ? = 1.5 ? and angle ? = 25∘. When it is released, it moves with an acceleration of 2 ? down the plane. a) Find the coefficient of kinetic friction ?? of the inclined plane. (??) b) Find the speed of Object 1 when it reaches the bottom of the plane. (??) At the bottom of the inclined plane, Object 1 arrives at a smooth (frictionless) horizontal surface where it hits Object 2 of three times mass of Object 1 (?2 = 3?1). c) Assuming an elastic collision, find the speeds of Objects 1 and 2 just after the collision

Answer 1

(a) Mass of object 1 be m₁ kg, and, g = 9.8 m / s² be the acceleration due to gravity.

Weight, m₁g, of object 1 has two components :

The component parallel to the incline : m₁g sin 25^o,

and, the compoent which supports the normal force : m₁g cos 25^o.

Net force along the incline = m₁ kg x 2 m / s².

Hence, frictional force = F_f = ( m₁g sin 25^o - 2m ) N = m₁ ( 9.8 sin 25^o - 2 ) N = 2.142m₁ N.

Hence, coeffitient of kinetic friction = _k = F_f / ( m₁g cos 25^o ) = 2.142m₁ / 8.88m₁

or, coeffitient of kinetic friction = _k = 0.24.

(b) Initial height of object 1 = h = d sin 25^o = 1.5 m x sin 25^o = 0.634 m.

Hence, initial potential energy = m₁gh = 6.2m₁.

Hence, its speed at the bottom of the incline u is vigen by : m₁u² / 2 = m₁gh = 6.2m₁

or, u = ( 2 x 6.2 ) = 3.52 m / s.

Hence, the speed of Object 1 when it reaches the bottom of the plane is : 3.52 m / s.

(c) For an elastic collision,

final speed of object 1 = v₁ = ( m₁ - m₂ ) u / ( m₁ + m₂ ) = ( - 2m₁ x 3.52 / 4m₁ ) m / s = - 1.76 m / s,

and. final speed of object 2 = v₁ = 2m₁u / ( m₁ + m₂ ) = ( 2m₁ x 3.52 / 4m₁ ) m / s = 1.76 m / s.