9. A bullet of mass 10g strikes a wooden plank at a speed of 500m/s and gets embedded after travelling 5cm. Calculate (i) the resistive force exerted by the wooden plank on the bullet (ii) the time taken by the bullet to come to rest.
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Answer:
A bullet with a mass of 20 g moving with a speed of 75 m/s hits a fixed wooden plank and comes to rest after penetrating a distance of 5 cm. What is the average resistive force exerted by the wooden plank on the bullet?
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v2−u2=2as
02−752=2a∗0.05
−5625=2a∗0.05
a=−56250m/s2
F=ma
F=0.02∗−56250=−1125N
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By the kinematics, we can find the retardation experienced by the bullet when it passes through the plank.
This can be done using the third kinematic equation v2−u2=2(a)(s)
since the bullet stops after penetrating, put v=0 m/s. the initial velocity with which bullet entered equals 75 m/s. the distance travelled equals 5/100 m. by substituting the values we get mag. a= 56250 m/s^2.
By Newton’s second law the avg. resistance or force experienced by bullet equals m(a). where m is mass of bullet. By substituting the values, we get mag. F= 1125N.
the mass and distance must be converted in to SI unit of system. Then you would get answer in newtons.