A bullet of mass 4 ~g is fired horizontally with a speed of 300 ~m / s into 0.8 ~kg block of wood at rest on a table. If the coefficient of friction between the block and the table is 0.3 , how far will the block slide approximately?

Step-by-step solution

Given, m ₁=4 ~g , u ₁=300 ~m / s m ₂=0.8 ~kg =800 ~g , u ₂=0 ~m / s From law of conservation of momentum, m ₁ u ₁+ m ₂ u ₂= m ₁ v ₁+ m ₂ v ₂ Let the velocity of combined system = v m / s then, 4 300+800 0=(800+4) v v = 1200 804 =1.49 ~m / s Now, =0.3 (given) aligned & a = g & a =0.3 10 ( take g =10 ~m / s ^2 ) &=3 ~m / s ^2 aligned then, from v ^2= u ^2+2 as (1.49)^2=0+2 3 s s = (1.49^2 ) 6 ; s = 2.22 6 =0.379 ~m