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

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

Step-by-step 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=√(0-2* 0.20* 9.8* 0.390)`

`=√(-1.5288)`

`=1.23 \ m/s`

As we know,

The conservation of momentum is:

⇒
`mu=m'u'`

or,

⇒ Initial speed,
`u'=(mu)/(m')`

On substituting the values, we get

⇒
`=(1.20* 1.23)/(0.009)`

⇒
`=(1.476)/(0.009)`

⇒
`=164 \ m/s`