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.

A bullet of mass 0.25kg is fired with velocity 302m//s into a block of

1.	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» Solution : (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`