A bullet of mass 20g is horizontally fired with a velocity of 150m/s from a pistol of mass 2 kg. What is the recoil velocity of the pistol?

m1 = 20g (= 0.02 kg)

Mass of pistol, m2 = 2 kg

Initial velocity of the bullet (u1) and pistol (u2) = 0

Final velocity of the bullet, v1 = +150m s-1

Let v be the recoil velocity of the pistol.

Total momentum of the pistol and bullet after it is fired is

= (0.02 kg x 150 m s-1) + (2 kg x v m s-1)

= (3 + 2v) kg m s-1

Total momentum after the fire = Total momentum before the fire

3 + 2v = 0

→v = -1.5 m/s  

prolly a beacuse the momentuem the car has increases while going down hill

another example of kinetic energy would be,

kinetic energy   >     moving energy

answer   >   letter choice (c),   a hammer falling off a table.

hope that !                                         : )