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  • A bullet of mass 0.012 kg and horizontal speed 70 m s1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block r

Q24 A bullet of mass 0.012 kg and horizontal speed 70 ms^{-1} strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Answers (1)

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We are given :                      

Mass of the bullet m: 0.012 Kg     

Mass of the block M: 0.4  Kg  

The initial velocity of the bullet u: 70  m/s 

The initial velocity of the block     :     0  

The final velocity of the system (bullet + block): v

For finding the final speed of system we will apply the law of conservation of momentum :

                                   mu_b\ +\ M(0)\ =\ (m\ +\ M)v

or                                             v\ =\ \frac{0.84}{0.412}\ =\ 2.04\ m/s

Now for the system, we will apply the law of conservation of energy :

                    The potential energy at the highest point  =   Kinetic energy at the lowest point

                                       (m+M)gh\ =\ \frac{1}{2}(m+M)v^2

or                                                         h\ =\ \frac{1}{2}\times \frac{v^2}{g}

or                                                               =\ \frac{1}{2}\times \frac{2.04^2}{9.8}

or                                                               =\ 0.2123\ m

Hence heat produced is :

                                          =\ \frac{1}{2}mu^2\ -\ \frac{1}{2}(m\ +\ M)v^2

or                                      =\ \frac{1}{2}(0.012)(70)^2\ -\ \frac{1}{2}(0.412)(2.04)^2

or                                      =\ 29.4\ -\ 0.857\ =\ 28.54\ J

Posted by

Devendra Khairwa

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