Question

A bullet of mass 0.04 kg moving with a speed of 90 m/s , enters a heavy wooden block and is stopped after transversing a distance of 6 cm . what is the average resistive force exerted by the block on the bullet

Answer 1

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Here's your answer...

Mass = 0.04 kg
Initial velocity u = 90 m/s
Final velocity v = 0 (as the bullet has come to rest)
Distance S = 0.06 m
By using the third formula of kinematics...
${v}^{2} = {u}^{2} - 2as \\ \\ {0}^{2} = {90}^{2} - 2 \times a \times 0.06 \\ 0.12a = 8100 \\ a = \frac{810000}{12} = 67500$
The retardation is 67500 m/s^2.

Force = mass * acceleration
Resistive force = mass * retardation
= 67500 * 0.04
= 2700 N

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