A bullet of mass 0.03 kg moving with a speed of 400 m/s penetrates 12 cm into fixed block.Calculate the average force exerted by the wood on bullet

Asked by RAJDEEP NASKAR | 18th Nov, 2013, 03:44: PM

Expert Answer:

Mass of the bullet (m) = 0.03 kg
Initial velocity of the bullet (u) = 400 m/s
Final velocity of the bullet (v) = 0 m/s
Distance travelled by the bullet = 12 cm
                                            = 0.12 m
 
Force (F) = ?
 
Force is given by,
 
F = ma                 ................(1)
 
v = u + at             ................(2)
 
Putting the given values in eq(2) we get,
 
0 = 400 + a×0.0003
 
a = - 1.3 × 106 m/sec2
 
From eq(1),
 
F = 0.03 kg × 1.3 × 106 m/sec2
 
F = 39 × 103 N = 39000 N
 
 

Answered by Komal Parmar | 18th Nov, 2013, 06:34: PM

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