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प्रश्न
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.
उत्तर
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
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