A bullet of mass `0.25kg` is fired with velocity `302m//s` into a block of wood of mass `m_1=37.5kg`. It gets embedded into it. The block `m_1` is resting on a long block `m_2` and the horizontal surface on which it is placed is smooth. The coefficient of friction between `m_1` and `m_2` is `0.5`. Find the displacement of `m_1` on `m_2` and the common velocity of `m_1` and `m_2`. Mass `m_2=1.25kg`.

Answer

A, D

Explanation

(i) Common velocity =("Initial momentum")/("Total mass")
`v_c=(0.25xx302)/(0.25+37.5+1.25)`
`=1.94m//s`
(ii) `v_1=("Initial momentum")/(m_1+m)`
`=(0.25xx302)/(37.5+0.25)=2m//s`
`a_1=(f)/(m_1+m)=mug=5m//s^2`
`a_2=(f)/(m_1+m+m_2)=((m_1+m)/(m_1+m+m_2))mug`
`=((37.5+0.25)/(37.5+0.25+1.25))(0.5xx10)`
`=4.84m//s^2`
`a_r=a_1-a_r=0.16m//s^2`
Common velocity is achieved when, `v_1` converts into `v_c` by a retardation `a_1`.
`:. v_c=v_1-a_1t`
`:. t=(v_1-v_c)/(a_1)=(2-1.94)/(5)`
`=0.012s`
Now, `s_r=1/2a_rt^2`
`=1/2xx0.16xx(0.012)^2`
`=0.011mm`