A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.390 m along the surface before stopping.

Required:
What was the initial speed of the bullet?

Respuesta :

Answer:

The initial speed of bullet is "164 m/s".

Explanation:

The given values are:

mass of bullet,

[tex]m'=9.00 \ g[/tex]

or,

    [tex]=0.009 \ kg[/tex]

mass of wooden block,

[tex]m=1.20 \ kg[/tex]

speed,

[tex]s=0.390 \ m[/tex]

Coefficient of kinetic friction,

[tex]\mu=0.20[/tex]

As we know,

The Kinematic equation is:

⇒  [tex]v^2=u^2+2as[/tex]

then,

Initial velocity will be:

⇒  [tex]u=v^2-2as[/tex]

        [tex]=v^2-2 \mu gs[/tex]

On substituting the given values, we get

⇒  [tex]u=\sqrt{0-2\times 0.20\times 9.8\times 0.390}[/tex]

       [tex]=\sqrt{-1.5288}[/tex]

       [tex]=1.23 \ m/s[/tex]

As we know,

The conservation of momentum is:

⇒  [tex]mu=m'u'[/tex]

or,

⇒ Initial speed, [tex]u'=\frac{mu}{m'}[/tex]

On substituting the values, we get

⇒                            [tex]=\frac{1.20\times 1.23}{0.009}[/tex]

⇒                            [tex]=\frac{1.476}{0.009}[/tex]

⇒                            [tex]=164 \ m/s[/tex]