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Q 14.     A bullet of mass 10 g travelling horizontally with a velocity of 150 m s-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

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By the equation of motion we know that  :- 

                                                   v\ =\ u\ +\ at

Putting values of v, u and t in the equation :

                                                   0\ =\ 150\ +\ a\times 0.03

or                                                a\ =\ \frac{-150}{0.03}

or                                               a\ =\ -5000\ m/s^2

Now we have :

                                         s\ =\ ut\ +\ \frac{1}{2}at^2

or                                            =\ 150\times 0.03\ +\ \frac{1}{2}\times (-5000)\times 0.03^2

or                                            =\ 2.25\ m

Hence the distance of penetration of the bullet in the block is 2.25 m.

Now for the force we have :

                                                  F\ =\ ma

or                                               =\ 0.01\times (-5000)\ =\ -50\ N

Thus the retarding force acting on the bullet is 50 N.


Posted by

Devendra Khairwa

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