Two wooden blocks of masses ` 1kg` and `2kg` are separated by a certain distance `A` bullet of mass `50g` fired from a gun pierces through the block of mass ` 1 kg` and then stopped in the second block After the impact of the bullet both blocks start moving with the same speed Calculate the percentage loss in the initial velocity of the bullet when it is inbetween the two blocks .

Explanation

Here, ` M_(1) = 1 kg M_(2) = 2 kg `
` m = 50 g = (50)/(1000) kg = (1)/(2) kg `
Let `upsilon` = initial velocity of the bullet
`upsilon_(1)` = velocity of the bullet after piercing through the first block
`V` = velocity of each block when hit by the bullet
As the bullet penetrates first block as per the principle of conservation of linear momentum ` m upsilon = M_(1) V + m upsilon_(1) `
Again applying the same principle when the bullet is embedded in second block :
` m upsilon_(1) = (M_(2) + m) V `
` upsilon_(1) = ((M _(2) + m ))/(m) = ((2 + 0.05))/(0.05) V = 41 V `
From (i) ` 0.05 upsilon = 1 V + 0.05 upsilon_(1) = V + 0.05 xx 41 V = 3.05 NV `
` upsilon = (3.05 V)/(0.05) = 61 V `
`%` age loss in initial velocity of bullet ` = ((upsilon - upsilon_(1) xx 100))/(upsilon) = ((61 V - 41V))/(61V) = (20)/(61) xx 100 = 32.8 % `
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