A bullet is fired into a fixed target looses one third of its velocity after travelling 4 cm . It penetrates further D × 10 - 3 m before coming to rest. The value of D is :

Step-by-step solution

When the bullet just enters the target, its equation of motion is given by v 2 - u 2 = 2 a S . . . 1 Equation (1) implies that 2 u 3 2 = u 2 + 2 - a 4 × 10 - 2 ⇒ 4 u 2 9 = u 2 - 2 a 4 × 10 - 2 ⇒ 5 u 2 9 = 2 a 4 × 10 - 2 … 2 When the bullet stops inside the target, its equation of motion can be written as 0 = 2 u 3 2 + 2 - a x . . . 3 Equation (3) implies that 4 u 2 9 = 2 a x . . . 4 Dividing equation (2) by equation (4), we have 5 4 = 4 × 10 - 2 x ⇒ x = 16 5 × 10 - 2 ⇒ x = 3 . 2 × 10 - 2 m = 32 × 10 - 3 m Hence, D