A4.1 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.

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

the answer is b

explanation:

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle