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Question
A bullet of mass 20g travelling with a velocity of 350ms-1 , comes to rest after penetrating 40cm in a still target. Find the resistive force offered by the still target and the retardation caused by it.
A bullet of mass 20g travelling with a velocity of 350ms-1 , comes to rest after penetrating 40cm in a still target. Find the resistive force offered by the still target and 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
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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