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Vadim26 [7]
3 years ago
14

A 9.00 g bullet is fired horizontally into a 1.20 kg wooden block resting on a horizontal surface. The coefficient of kinetic fr

iction 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?
Physics
1 answer:
Zinaida [17]3 years ago
3 0

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                              

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