A bullet of mass 10g moving horizontally at a speed of 50 7 ​ m/s strikes a block of mass 490g kept on a frictionless track as shown in figure. The bullet. remains inside the block and the system proceeds towards the semicircular track of radius 0.2m. 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`