A bullet of mass 50g moving with velocity `600m//s` hits block of mass `1.0kg` placed on a rough horizontal ground and comes out of the block with a velocity of `400m//s`. The coefficient of friction between the block and the ground is `0.25`. Neglect loss of mass of the block as the bullet pierces through it.
(a) In spite of the fact that friction acts a an external force, can you apply principle of conservation of momentum during interaction of the ullet with the block?
(b) Find velocity of the block immediately after the bullet pierces through it. (c)FInd the distance the block will travel before it stops.

Explanation

(a) There is no net external force in the verticle direction and in the horizontal direction, only external force friction is non-impulsive, therefore momentum of the bullet-block system during their interaction remains conserved.
(b) Let us denote velocities of the bullet before it hits the block and immediately after it oierces through the block by `v_(b0)` and `v_(b)`, velocity of the block immediately after the bullet pierces through it is `v_B` and masses of the bullet and the block by m and M respectively. These are shown in the adjacent figure.Applying principle of momentum for horizontal component, we have `mv_(b0) = mv_(b) + Mv_(B) rarr v_(B) = (m(v_(b0
))-V_(b))/M`
Substituting the given values, we have `v_(B) = 10m//s`

(c) To calculate distance traveled by the block before it stops, work kinetic energy theorem has to be applied.
DUring sliding of the box on the ground only the force of kinetic friction does work.
`W_(1rarr2) = K_2-K_1rarr " " -muMgx=0-1/2Mv_B^(2)`
`x = (v_(B)^(2))/(2mug)`
Substituting given values, we have `x=20m`