Question of Physics

Question

A bullet of mass m hits a block of mass M on a table with velocity v₀. The bullet embeds in the block. The friction coeffcient between the block and the table is µ = µ₀.x where µ₀ is a constant and 0 ≤ x ≤ d. Find the distance d taken by the block before it stops.

Answer 1

Let g be gravitational acceleration.

After the collision, the block with the bullet imbedded has weight (m+M)g,

and the force of friction is then (m+M)gμ = (m+M)gμ₀x.

The initial kinetic energy, mv₀²/2, of the bullet gets converted to the work of friction, which is

∫ [0, d] (m+M)gμ₀xdx = (m+M)gμ₀d²/2

By conservation of energy:

mv₀²/2 = (m+M)gμ₀d²/2

d = v₀√(m/((m+M)gμ₀)))

You might notice that the units appear to be wrong;

the units of the result should be 'length', i.e., m, but are

m/s √(s²/m) = √(m)

The reason for the discrepancy is that the coefficient of friction,

which is a dimension-less quantity is given to us as if it had dimension of length.

So the numerical result we got for d is actually in m, units of length.