A lead bullet of mass 20 g, travelling with a velocity of 350 m s-1, comes to rest after penetrating 40 cm in a still target. Find (i)the resistive force offered by the target (ii)the retardation caused by it.

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Solution

Mass of the bullet, m = 20 g = 0.02 kg
Initial velocity of the bullet, u = 350 m/s
Final velocity of the bullet, v = 0
Distance traveled, S = 40 cm = 0.4 m
Using, v² = u² + 2aS
=> 0 = 350² + 2(a)(0.4)
=> a = -153125 m/s²
The negative sign comes because acceleration opposes the motion.
So, retardation is = 153125 m/s²
Thus, the resistive force is = (.02)(153125) = 3062.5 N

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