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A 4.1 g bullet is fired horizontally at a block of mass 0.90 kg resting on a frictionless tabletop. The bullet becomes embedded in the block and a speed of 4.0 m/s is imparted to the block. Find the initial speed of the bullet before it hits the block.

Respuesta :

Momentum before collision:

= (Mass of bullet x velocity of bullet) + (Mass of block x velocity of block)

= (4.1 Kg/1000 x V m/s) + ( 0.90 kg x 0)

=  Momentum after collision

Momentum after collision:

= (Mass of bullet + Mass of block ) x velocity of bullet

= (4.1 Kg/1000 + 0.90 Kg) 4.0 m/s

= 3.6164 Kg-m/s

Momentum before collision = Momentum after collision

4.1 Kg/1000 x V m/s = 3.6164 Kg-m/s

V = 882.048 m/s

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