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.

Question

Physics

Kimora Donaldson

6 May, 02:23

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.

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Nia Lamb · Answer 1

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