Question

A block of mass m is moving on rough horizontal surface with momentum p. The coefficient of friction between the block and surface is µ. The distance covered by the block before it stops is [g =acceleration due to gravity)

Options

• `(2mu"M"^2"g")/"p"^2`

• `"p"/(2mu"Mg")`

• `(2mu"Mg")/"p"`

• `"p"^2/(2mu"M"^2"g")`

Answer 1

`"p"^2/(2mu"M"^2"g")`

Explanation:

Using kinematic relation,

v² = u² - 2as

The momentum pis given by

p = Mu

u = `"P"/"M"`

and acceleration is given as

a = μg

Substituting values in Eq. (i), we get

0 = u2 - 2as  `(because final velocity, v = 0)`

0 = `("p"/"M")^2 - 2mu"gs"`

`("p"/"M")^2 = 2mu"gs"`

`=> "s" = "p"^2/(2"M"^2mu"g")`