Question

Help thanks?!

Answer 1

See below.

Explanation:

Equating the process we have

`m_1 v_1+m_2 v_2 = (m_1+m_2)v` Momentum

`1/2(m_1+m_2)v^2=mu (m_1+m_2)g d` Energy transmitted to the wooden block.

`E_d = mu (m_1+m_2)g d` Movement dissipated energy

`E_c = 1/2m_1v_1^2-E_d` Collision lost energy

Here

#{(),(m_1 = 0.004 ),(m_2 = 800),(v_1= 300),(v_2=0),(g=9.81),(mu=0.3),(d = 3.8 xx 10^-7 " distance moved by the block"),(v = 0.0015),(E_d=0.0009 "),(E_c approx 180):}#

The results are poor. Perhaps due to improper values to the bullet mass or the bullet speed.

Answer 2

It is assumed that the bullet gets embedded in the wooden block after collision.

(a) During collision momentum is conserved.
Let the Block`+`Bullet combo move with a velocity `v` after collision.

`"Initial momentum"="Final momentum"`
`0.004xx300=800.004xxv`
`=>v=(1.2)/800.004=0.0015ms^-1`

Kinetic energy of combo `=p^2/(2m)=(0.004xx300)^2/(2xx800.004)=0.0009J`

Force of friction `=mumg=0.3xx800.004xx9.81=2354.4N`

If `d` is distance moved by the combo, work done by force of friction is

`W=Fxxd`.

This work done must be equal to kinetic energy of the combo. Equating both we get

`2354.4d=0.0009`
`=>d=0.00000038m`

(b) Kinetic energy of bullet `=1/2xx0.004xx(300)^2=180J`
Kinetic energy of bullet lost in collision `=(180-0.0009)`
Fraction of bullet's energy lost in collision `=(180-0.0009)/180~~1`