Question

A little boy throws a 100g lump of clay horizontally at a 700g box sitting on a table. The lump of clay hits the box at the velocity of 25m/s and sticks to it, causing the box to slide 1.5m across the table before stopping. What is the kinetic friction coefficient µk between the box and the table?

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the lump of clay = m = 100 g = 0.1 kg

Mass of the box = M = 700 g = 0.7 kg

Velocity at which the clay hits the box = V₁ = 25 m/s

Initial velocity of the box = V₂ = 0 m/s

Velocity of the box and lump of clay after the collision = V₃

By conservation of linear momentum,

mV₁ + MV₂ = (m + M)V₃

(0.1)(25) + (0.7)(0) = (0.1 + 0.7)V₃

V₃ = 3.125 m/s

Normal force on the box from the table = N

N = (m + M)g

Coefficient of kinetic friction between the box and the table = _k

Distance the box slides before stopping = d = 1.5 m

The kinetic energy of the box and clay after the collision is lost to work done against friction as the box moves across the table.

_k = 0.332

Kinetic friction coefficient between the box and the table = 0.332