Question

The 10-kg block is held at rest on the smooth inclined plane by the stop block at A. If the10-g bullet is traveling at 300m/swhen it becomes embedded in the 10-kg block, determine thedistance the block will slide up along the plane before momentarily stopping

Answer 1

6.8 mm

Step-by-step explanation:

We are given that

Mass of block,m=10 kg

Mass of bullet,
`m_b=10 g=10* 10^(-3) kg`

1 kg=1000 g

Total mass of system,M=
`m+m_a=10+10* 10^(-3)=10.01kg`

Speed of bullet,u=300 m/s

`\theta=30^(\circ)`

By law of conservation of momentum

`m_bucos\theta=Mv`

`v=(m_bvcos\theta)/(M)=(0.01* 300cos30^(\circ))/(10.01)=0.259m/s`

According to law of conservation of energy

Change in kinetic energy of system=Change in potential energy of system

`(1)/(2)Mv^2-0=Mgh-0`

`(1)/(2)(10.01)(0.259)^2=10.01* 9.8 h`

Where
`g=9.8 m/s^2`

`h=((0.259)^2)/(2* 9.8)=0.0034m`

1m=100 cm

`h=0.0034* 100=0.34 cm`

Distance traveled by block=
`d=(h)/(sin\theta)=(0.34)/(sin30^(\circ))=0.68 cm=6.8 mm`

1cm=10 mm