Download Ch 9--Linear Momentum and Collisions #1

Lab: AP Review Sheet
Chapter 9: Linear Momentum and Collisions
AP Physics
Erynn Kim
Background: The concept of momentum is useful for describing objects in motion. The concept of
center of mass can be used to describe the motion of a system of particles through the motion of one
representative particle located at the center of mass.
Major Topics:
• Linear Momentum
• Collisions (Inelastic and
Elastic) in One Dimension
and in Two Dimensions
• The Center of Mass
Formulae:


p ≡ mv




p1i + p2i = p1 f + p2 f

Ft = Δp
dp
∑ F = dt
J=
∫
tf
ti
F dt = Δp

mi ri
∑

rcm =
M
1 

rcm =
∫ r dm
M
Vocabulary:
definition of linear momentum of a particle: the linear
momentum of a particle or an object that can be modeled as a

particle of mass m moving with a velocity v is defined to be the
product of the mass and velocity of the particle.
Newton’s second law for a particle: the time rate of change of the
linear momentum particle is equal to the net force acting on the
€
particle.
conservation of momentum (isolated system): whenever two or
more particles in an isolated system interact, the total momentum of
the system remains constant.
collision: an event during which two particles come close to each
other and interact by means of forces
elastic collision: a collision in which the total kinetic energy and
total momentum of the system is the same before and after the
collision
inelastic collision: collision in which the total kinetic energy of the
system is not the same before and after the collision
perfectly inelastic collision: inelastic collision in which the objects
stick together after they collide
center of mass: the point in a system around which the mass of the
system is distributed
€
Inelastic
Elastic collisions: K1 + K 2 = K1 '+K 2 ' & p1 + p2 = p1 '+ p2 '
1
1
1
1
m1v12 + m2v 2 2 = m1v1 '2 + m2v 2 '2 & m1v1 + m2v 2 = m1v1 '+m2v 2 '
2
2
2
2
€
€
€
€
Inelastic collisions: p1 + p2 = p1 '+ p2 ' or m1v1 + m2v 2 = m1v1 '+m2v 2 '
€
Perfectly inelastic collisions:
p1 + p2 = p€
1 '+ p2 ' or m1v1 + m 2v 2€= ( m1 + m 2 )v 2 '
€
Elastic
Problems:
1. [Easy]
Four objects are situated along the x-axis as follows: a 3.00-kg object is at 4.00 m, a 2.00-kg object is
at +1.50-m, a 1.50-kg object is at the origin, and a 4.00-kg is at -0.750 m. Where is the center of
mass of these objects?
SOLUTION:
4.00-kg, -0.750 m
2.00-kg, 1.50 m
1.50-kg, 0 m
0
-1
xCM
€
1
2
3
4
m1 x1 + m2 x 2 + m3 x 3 + m4 x 4
m1 + m2 + m3 + m4
(3.00kg)(4.00m) + (2.00kg)(1.50m) + (1.50kg)(0m) + (4.00kg)(−.750m)
=
3.00kg + 2.00kg +1.50kg + 4.00kg
= 1.14m
xCM =
xCM
3.00-kg, 4.00 m
Problems (cont’d)
2. [Medium]
An unstable atomic nucleus of mass 18.0x10-27 kg initially at rest disintegrates into three particles.
One of the particles of mass 6.00x10-27 kg, moves in the y direction with a speed of 7.00x106 m/s.
Another particle, of mass 8.50x10-27 kg, moves in the x direction with a speed of 5.00x106 m/s. Find
(a) the velocity of the third particle and (b) the total kinetic energy increase in the process.
m1 = 6.00x10-27 kg
v1 =7.00x106 m/s SOLUTION:
m = 18.0x10-27 kg
v =0 m/s m2 = 8.50x10-27 kg
v2 =5.00x106 m/s m3 = ?
v3 =? Original
Final
a. Conservation of momentum (total final momentum of system is m1v1 f + m2v 2 f + m3v 3 f
and it must be zero to equal the original momentum.
The mass of the third particle is:
m3 = (18.0 − 6.00 − 8.50) × 10 −27 kg
€
m3 = 3.50 × 10 −27 kg
Solving:
m1v1 f + m2v 2 f + m3v 3 f = 0
€
v3 f = −
(6.00 × 10 −27 kg) (7.00 × 10 6 ˆj )m /s + (8.50 × 10 −27 kg) (5.00 × 10 6 iˆ )m /s
(
)
3.50 × 10
(
−27
)
kg
(4.20 ˆj + 42.5iˆ ) × 10 kg⋅ m /s
3.50 × 10 −27 kg
= (−1.21iˆ −1.20 ˆj ) × 10 7 m /s
−20
v3 f = −
v 3 f
b. Original kinetic energy of system is 0.
€
Final kinetic energy is:
K = K1 f + K 2 f + K 3 f
1
K1 f = (6.00 × 10 −27 kg)(7.00 × 10 6 m /s) 2 = 1.47 × 10 −13 J
2
1
K 2 f = (8.50 × 10 −27 kg)(5.00 × 10 6 m /s) 2 = 1.06 × 10 −13 J
2
1
K 3 f = (3.50 × 10 −27 kg) × [(−1.21 × 10 7 m /s) 2 + (−1.20 × 10 7 m /s) 2 ] = 5.08 × 10 −13 J
2
€
€
€
So the system kinetic energy is:
K = 1.47 × 10 −13 J +1.06 × 10 −13 J + 5.08 × 10 −13 J
K = 7.61 × 10 −13 J
Problems (cont’d)
3. [Hard]
A 4.00-g bullet moving with an initial speed of vi=375m/s is fired into
and passes through a 0.75-kg block. The block, initially at rest on a
frictionless, horizontal surface, is connected to a spring with a force
constant 800 N/m. The block moves d=4.00 cm to the right after
impact before being brought to rest by the spring. Find (a) the speed
at which the bullet merges from the block and (b) the amount of initial
kinetic energy of the bullet that is converted into internal energy in the
bullet-block system during the collision.
SOLUTION:
Need to find initial velocity of block first (conservation of energy during spring compression).
(K + U ) before = (K + U ) after
1
1
m2v B 2 = kx 2
2
2
k
vB =
x
m2
vB =
800N /m
0.0400m
0.75kg
v B = 1.31m /s
a. After collision, momentum is conserved.
m1v1i + m2v 2i = m1v1 f + m2v B
€
v1 f = ( m1v1i − m2v B ) /m1
v1 f
(4.00 × 10
=
−3
kg)(375m /s) − (0.75kg)(1.31m /s)
4.00 × 10 −3 kg
v1 f = 129m
/s
€
€
b. Need to think about energy in isolated system to find the mechanical energy converted into
internal energy in the collision. Before collision, block is motionless, bullet’s energy is:
1
1
K i = m1v1i2 = (0.00400kg)(375m /s) 2 = 281J
2
2
After collision, mechanical energy is:
1
1
K f = m1v12f + m2v B2
2
2
1
1
K f = (0.00400kg)(129m /s) 2 + (0.75kg)(1.31m /s) 2 = 33.9J
2
2
So energy converted from mechanical to internal in collision is given by ΔK + ΔE int = 0 :
ΔE int = −ΔK = −K f + K i = −33.9J + 281J = 247J
€
€
€