Question

A bullet of mass 5g travelling at a speed of 120m/s penetrates deeply into a fired target and is brought to rest in .01 sec.Calculate
a) the distance of penetration in the target
b) average force exerted on the bullet

Answer 1

u = 120 m/s
v = 0
t = 0.01 s

a = (v-u)/ t = -120 / 0.01 = -12000 m/s

a) S = (v² - u²)/2a = -120² / 2*(-12000) = 120 / 200 = 0.6 m
distance of penetration is 0.6m

b) force = ma = 0.005×12000 = 60N

Answer 2

Answer:

a) the distance of penetration in the target= 0.6m

b) average force exerted on the bullet = - 60 N

Explanation:

• Mass of the bullet, m = 5g = 0.005 kg
• Initial speed. u = 120 m/s.
• After penetrating into the target the speed of bullet will decrease and finally its speed will become 0. So v = 0.
• Time taken, t = 0.01s.
• Acceleration, a = (v-u)/t = (0-120)/0.01 = - 12000 m/s²
• Suppose the distance covered is 's'.
• From the 3rd equation of motion, v² - u² = 2as

                                                            ⇒ s = (v² - u²)/2a

                                                                   = (0 - 120²)/2*(-12000)

                                                                   = 120*120/2*12000

                                                                   = 6/10 = 0.6m

        ∴The distance of penetration in the target = 0.6m.

• Force = ma = 0.005 x (- 12000) N

                            = - 5 x 12 N

                            = - 60 N

          ∴The average force exerted on the bullet = - 60 N .