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DIA [1.3K]
3 years ago
6

A 7.0- g bullet is fired into a 1.5- kg ballistic pendulum.The

Physics
1 answer:
777dan777 [17]3 years ago
4 0

Answer:

the initial velocity of bullet is u = 330.335 m /s

Explanation:

given,

mass of bullet(m) = 7 g

Mass of pendulum ( M ) = 1.5- kg

The bullet emerges from the block with a speed (V) = 200 m/s

The maximum height ( h ) =   12 cm.

Let the initial speed of the bullet be ' u '

Apply, Law of conservation of momentum :

           m u = ( m + M ) V  ...............................(1)

Apply, Law of conservation of Energy :

         \dfrac{1}{2}( m + M ) V^2 = ( m + M )g h  ...............(2)

From equation (1) and  equation (2) , we have

                            u = \dfrac{m + M}{m}V  

But V = √2gh

                   u=\dfrac{m + M}{m}\times \sqrt{2gh}  

                   u=\dfrac{0.007+1.5}{0.007}\times \sqrt{2\times 9.81\times 0.12}


                          u = 330.335 m /s

hence, the initial velocity of bullet is u = 330.335 m /s

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Answer:

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for this problem we have to convert the time interval ins seconds, we know that a year has 53926560s

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then we can use the ecuation number 1 to calculate the aceleration

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