Two wooden blocks of masses1kg and 2kg are separated by a certain distance A bullet of mass 50g fired from a gun pierces through the block of mass1 kg and then stopped in the second block After the impact of the bullet both blocks start moving with the same speedCalculate the percentage loss in the initial velocity of the bullet when it is inbetween the two blocks .

Two wooden blocks of masses1kg and 2kg are separated by a certain dist

1.	Two wooden blocks of masses1kg and 2kg are separated by a certain distance A bullet of mass 50g fired from a gun pierces through the block of mass1 kg and then stopped in the second block After the impact of the bullet both blocks start moving with the same speedCalculate the percentage loss in the initial velocity of the bullet when it is inbetween the two blocks .
Answer» Solution :Here, ` M_(1) = 1 kgM_(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 eachblock 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 principlewhen 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 % ` .