Question

Bullet of 10 gram strike send bag at 10th cube metre per second and get embed at her travelling 5 cm

Answer 1

the depth of bullet = 5cm = 0.05cm mass of bullet = 10gm = 0.01 kg now by third eqn. of motion v^2 – u^2 = 2as 0 – 1000×1000 = 2 × a ×0.05 a = –10000000 m/s^2 here -ve sign indicates that acceleration is opposite to the motion. now a). force = ma  = 0.01 × 10000000 = –100000 N  b). time taken by the bullet to get stopped = v = u + at => 0 = 1000 – 100000×t 1000 = 100000 × t =>t = 0.01 sec