Question

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?

Answer 1

Answer:

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

Explanation:

The given values are:

mass of bullet,

$m'=9.00 \ g$

or,

    $=0.009 \ kg$

mass of wooden block,

$m=1.20 \ kg$

speed,

$s=0.390 \ m$

Coefficient of kinetic friction,

$\mu=0.20$

As we know,

The Kinematic equation is:

⇒  $v^2=u^2+2as$

then,

Initial velocity will be:

⇒  $u=v^2-2as$

        $=v^2-2 \mu gs$

On substituting the given values, we get

⇒  $u=\sqrt{0-2\times 0.20\times 9.8\times 0.390}$

       $=\sqrt{-1.5288}$

       $=1.23 \ m/s$

As we know,

The conservation of momentum is:

⇒  $mu=m'u'$

or,

⇒ Initial speed, $u'=\frac{mu}{m'}$

On substituting the values, we get

⇒                            $=\frac{1.20\times 1.23}{0.009}$

⇒                            $=\frac{1.476}{0.009}$

⇒                            $=164 \ m/s$