Question

Two wooden plank of mass $M_1=1kg,M_2=2.98kg$ having smooth surface. A bullet of mass $m=20$ gm strikes the block $M_1$ and pierces through it, then strikes the plank B and sticks to its. Consequently both the planks move with equal velocities. If the percentage change in speed of the bullet when it escapes from the first block is x/3 %, find the value of x.

Answer 1

Given : Mass of bullet $m=0.02kg$

$M_1=1kg$ $M_2=2.98kg$

Let the velocity of bullet before and after first collision be $u$ and $v^{\prime }$ respectively.

Also let the velocity of First plank be $V$ after Ist collision and that of " Second plank + bullet" system be $V$ after second collision.

Applying conservation of momentum at initial and after Ist collision :

$mu=m{v}^{\prime }+M_{1}V$

$\left(0.02\right)u=\left(0.02\right)v^{\prime }+V$ ...............(1)

Applying conservation of momentum at initial and after 2nd collision :

$mu=(M_2+m)V$

$\left(0.02\right)u=(2.98+0.02)V$ $⟹V=\frac{0.02}{3}u$

From equation (1), $0.02u=0.02v^{\prime }+\frac{0.02}{3}u$

$⟹$ $v^{\prime }=\frac{2}{3}u$

Thus fraction change in speed of bullet due to Ist collision $\frac{\Delta v}{u}=\frac{u-\frac{2}{3}u}{u}=\frac{1}{3}$

$\therefore$ Percentage change in speed of bullet $=\frac{\Delta v}{u}\times 100=\frac{100}{3}$

$⟹x=100$