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The Law of Conservation of Momentum

Worksheet

Jerish Mae Cortado

United States of America

DEFAULT

Physics

Law of conservation of momentum

Introduction to momentum

Conservation of momentum

The Law of Conservation of Momentum
The law of conservation of momentum says that, "If the total external force on a system is equal to
zero, then the final value of the momentum of the system is equal to the initial value of the total
momentum of the system."
Thus,
Solution:
From conservation of momentum,
The equation above illustrates the law of conservation of momentum. That is, the total momentum is
constant, or usually stated, the total momentum is conserved.
=
Collisions
F = = 0
Ap
At
Example
A 15-g bullet is fired from a 5-kg rifle at a muzzle velocity of 600 m/s. Find the recoil velocity of the rifle.
Solution:
From conservation of momentum, mv, m₂vb₁
Ap
At
=
= 0
P2 = P1
Wa
mb
Vr = (vb)
\mr
0.015 kg
5 kg
Example
An astronaut is in space at rest relative to an orbiting spacecraft. His total weight is 300 lb, and he
throws away a 1-lb wrench at a velocity of 15 ft/s relative to spacecraft. How fast does he moves off in
the opposite direction?
(600 m/s)
= 1.8 m/s
mava = mwvw
) (v₁) = (W) (vw)
va = (W) (P₂)
Va
(vw)
Wa
1 lb
(15 ft/s)
300 lb
= 0.05 ft/s Momentum is also conserved in collision. If a moving billiard ball strikes a stationary one, the two move
off in such a way that the vector sum of their momenta is the same as the initial momentum of the first
ball. This is true even if the balls move off in different directions.
UA
Before
U₂ = 0
m
VA
KEint = mv²
V
Inelastic Collision
A collision in which the total kinetic energy after the collision is less than that before the collision is
called an inelastic collision. In one kind of inelastic collision the colliding bodies stick together and move
as a unit after the collision.
P₁ + P₂ = 0
(a)
System of interest
B
-V
Frictionless surface
VB
net F = 0
V₂
m
A
After
B
System of interest
KEint = 0
p₁ + P₂ = 0
m
m
(b)
V = 0
Frictionless surface Example
Suppose that the collision in the figure below, the two bodies stick together after the collision; mA = 0.3
kg, VA = 2 m/s, MB = 0.5 kg and VB = - 2 m/s. Find the final velocity, and compare the initial and final
kinetic energies.
VA₁ = 2 m/s
VB₂ = -2 m/s
A
Solution:
Momentum before collision = momentum after collision
MAVA ₁ + MBVB₁ = MAVA₂ + MB VB₂
Since after the collision the two bodies were stuck together, then VA₂ = VB₂ = V.
MAVA₁ + MBVB₁ = (mA + MB)v
MAVA₁ + MBVB₁
MA + MB
The initial kinetic energy is
B
The final kinetic energy is
Lost energy is
V =
(0.3 kg) (2 m/s) + (0.5 kg)(-2 m/s)
0.3 kg + 0.5 kg
= -0.5 m/s
1
KE₁ = MAV₁, +MBV²₁
=(0.3 kg)(2 m/s)² + (0.5 kg) (-2)= = 1.6 J
1
KE₂ = (m₁ + MB) v²
=(0.3 kg +0.5 kg)(0.5 m/s)² = 0.10 J
Elost = KE1 - KE₂ = 1.6 J-0.10 J = 1.50 J Percentage of lost energy
Vli
% KElost =
(a) Before the collision
VA1 = 2 m/s
1.6 J-0.10J
1.6 J
Elastic Collision
An elastic collision is one in which the bodies involved move apart in such a way that kinetic energy as
well as momentum is conserved.
A
KE₁ - KE₂
KE₁
mA = 0.50 kg
x 100%
x 100% = 93.75 %
Solution:
Momentum before collision = momentum after collision
vysin 0
Example 13.
Suppose the collision shown in the figure below is elastic. What are the velocities of A and B after the
collision?
VB1 = -2 m/s
B
0
-Vasin o
(b) After the collision
MB = 0.30 kg
Vi
v cos
Vaj Cos
V2f Conservation of energy
Substitute the value of VA₂
From the quadratic formula
KE₁ = KE₂
1
1
1
1
mav², += mgv³₂ = {mav²₂ + ½mBV³₂
MAV²₁ + m₂v₁ = MAV²₂ + M³V³₂
(0.5) (2)² + (0.3) (-2)² = 0.5v²₂ + 0.3v³/2₂
0.5v²₂ +0.3v₂ = 3.2
V²₂ +0.6v₂ = 6.4
VB2
MAVA₁ + MBVB₁ = MAVA₂ + MBVB₂
(0.5) (2) + (0.3) (-2) = 0.5vA₂ + 0.3VB₂
0.5vA₂ + 0.3VB₂ = 0.4
VA₂+0.6VB₂ = 0.8
VA₂ = 0.8 -0.6VB₂ equation 1
Then the velocity of block A is
(0.8 -0.6vB₂)² + 0.6v² = 6.4
0.64 -0.96vB₂ +0.36v³₂ + 0.6v³₂ = 6.4
0.96v32 -0.96VB₂ + 0.64 - 6.4 = 0
0.96v3₂ -0.96VB₂ - 5.76 = 0
X =
-b ± √b² - 4ac
2a
-(-0.96) ± √(-0.96)²4(0.96)(-5.76)
Based on the figure, after the collision block B will moved to the right, so that its final velocity is
VB₂
2(0.96)
0.96±4.8
2(0.96)
=
0.96 + 4.8
2(0.96)
= 3 m/s
VA₂ = 0.8 -0.6VB₂
= 0.8 0.6(3 m/s) = -1 m/s

Law of conservation of momentum

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Law of conservation of momentum

Similar Content

Can't find what you're looking for? Explore other subjects.

Knowunity is the # 1 ranked education app in five European countries

Knowunity was a featured story by Apple and has consistently topped the app store charts within the education category in Germany, Italy, Poland, Switzerland and United Kingdom. Join Knowunity today and help millions of students around the world.

4.9+

13 M

#1

950 K+

Still not sure? Look at what your fellow peers are saying...

I love this app so much [...] I recommend Knowunity to everyone!!! I went from a C to an A with it :D

The application is very simple and well designed. So far I have found what I was looking for :D

Love this App ❤️, I use it basically all the time whenever I'm studying