Question

Using the second law of motion, derive the relation between force and acceleration. A bullet of 10 g strikes a sand-bag at a speed of 103 m s-1 and gets embedded after travelling 5 cm. Calculate

(i) the resistive force exerted by the sand on the bullet

(ii) the time is taken by the bullet to come to rest.

Answer 1

Rate of change of momentum = `("Change in momentum")/("Time taken")`

= `("m"("v" - "u"))/"t"`  ...........(i)

using v = u + at

∴ a = `("v" - "u")/"t"`  .............(ii)

From (i) and (ii), we get

Rate of change of momentum = ma

According to newton's second law of motion

Force ∝ rate of change of momentum

F ∝ m × a

Then F = K ma; K is a constant

∴ F = ma

Here mass of bullet, m = 10 g = 0.01 kg

Initial speed u = 10³ ms⁻¹

Final peed v = 0

Distance travelled, s = 5 cm = 0.05 m

(i) Using v² − u² = 2as, we get

0 − (10³)² = 2 × a × 0.05 = 0.1a

or a = `(-10^6)/0.1 = -10^7 "ms^-2"`

Therefore, the resistive force exerted by the sand on the bullet = − (Force exerted by bullet on sand)

= −ma = − 0.01 × (10⁷) N = +10⁵ N

(ii) Time taken by the bullet to come to rest,

t = `("v" - "u')/"a"`

= `(0 - 10^3)/-10^7`

s = 10⁻⁴ s