A bullet of mass 10 g moving horizontally at asped dof `50sqrt7m/s` strikes a block of mass 490 g kept on a frictionless track as shown in figure. The bullet remains inside the block and the system proceeds towards the semicircular tract o radius 0.2 m. Where will the block strike the horizontal part after leaving the semicircular track?

Explanation

mass of block =490 gm
mss of bullet = 10gm
Since the bullet embedded inside the block, it is a plastic collision
Initial velocity of bullet
`v_1=50 sqrt7m/s`
velocity of the block is `v_2=0`
Let fiN/Al velocity of boty =v
Hence `10xx10^-350 sqrt7+490xx10^-3xx0`
`(490+10)`
`=10^-3xxV_A`
`:. V_A=sqrt7m/s` ltbr. When the block loses the contat at D the component mg will act on it,
`(m(V_B)^2)/r=mgsintheta`
`rarr (V_B)^2=grsintheta` .........i
puting work energy principle
`(1/2)mxx(V_B)^2-(1/2)xxmxx(V_A)^2`
`=-mg(0.2+0.2sintheta)`
`rarr(1/2)xxgrsintehta-(1/2)xx(sqrt7)^2`
`=g(0.2+0.2sintheta)`
`rarr 3.5-(1/2)+9.8+0.2+sintheta`
`=9.8+0.2(1+sintheta)`
rarr 3.5-0.98sinthea=1.96+1.96sintheta`
`rarr sintheta=(1/2)`
`rarr theta=30^0`

`:. Angle of projection `=90^0-30^0=60^0`
`:.` time of reaching the ground `=(sqrt(2h)/g)`
,` =(sqrt(2x(0.2+0.2xxsin30^0)/9.8)=0.247sec`
`rarr Distance travelled in horizontal direction
`S=vcos q xxt`
`=sqrt(grsintheta) cos thetaxxt `
`=sqrt(9.8xx2xx(1/2)sqrt3/2xx0.247`
`0.196m`
Total distance `=(0.2-0.2cos30^0+0.196)`
`=0.22m`