Conservation of Momentum
If two isolated objects interact (collide
or separate), then the total momentum
of the system is conserved (constant).
pi=pf
M (fuel)V (fuel) + M (rocket)V (rocket) = M (fuel)Vʹ
0 = M (fuel)
Vʹ(fuel)
(fuel)
+ M(r)Vʹ(r)
0 + 0 = M (fuel)Vʹ(fuel) + M(r)Vʹ(r)
+
M(r) ʹ
V
(r)
Where Vʹ (fuel) [-V] is opposite Vʹ (rocket
M (fuel)
Vʹ(fuel)
=
M(r) ʹ
V
(r)
A bullet of mass 20 g is horizontally fired with a velocity 150 m/s from a pistol
of mass 2 kg. What is the recoil velocity of the pistol?
mass of bullet (m1)= 20 g = 0.02 kg
mass of the pistol (m2)= 2 kg
initial velocities: bullet (vb) and pistol (vp) = 0
The final velocity of the bullet ( vb)= + 150 m/s.
According to the law of conservation of momentum….
pi=pf
Total momenta of the pistol and bullet before the fire= 0
Total momenta of the pistol and bullet after it is fired
= [0.02 kg × (150 m/s)]+ [2 kg × v’p]
..OR…
0= (3 kg m/s)+ 2v’p …..
…OR….
v = − 1.5 m/s.
Negative sign indicates that the direction
in which the pistol would recoil is
opposite to that of bullet, that is, right to left.
Watermelon 2:04
A common pool shot involves hitting a ball into a pocket from an
angle. Shown below, the cue ball hits a stationary ball at an angle
of 60 o , such that it goes into the corner pocket. Both balls have a
mass of .5 kg, and the cue ball is traveling at 4 m/s before the
collision. What is the velocity of the cue ball after the collision?
60⁰
30⁰
p= 2 Ns
60⁰
30⁰
p= 2 Ns
p i=pf
2 Ns
p = mv
1.7 Ns=(0.5 kg)v
v c =3.4 m/s
@ -30⁰