Question

10. A bullet of mass 2 g fired from a gun acquired velocity of 40 m/s. It is then stopped by the target
and it travels 20 cm deep in the target. Calculate : [2]
a. Retardation produced
b. Retarding force.

Answer 1

Answer:

a. initial velocity=40m/s

final velocity=0m/s

distance=20cm=0.2m

according to the third equation of motion,

v²=u²-2as

0= 40²-2(0.2)a

0=1600-0.4a

-1600= -0.4a

a=4000m/s²

b. f=ma

f= 2×4000

f=8000N

Explanation:

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