A 1500 kg car traveling at 20 m/s collides with a 1000 kg car at rest. After the collision, they stick together. Find: (a) the final velocity, (b) the initial kinetic energy, (c) the final kinetic energy, and (d) the energy lost.
Conservation of momentum, elastic and inelastic collisions
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Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 3 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Momentum and Collisions is part of the AP Physics C: Mechanics course on Study Mondo, specifically in the Linear Momentum section. You can explore the full course for more related topics and practice resources.
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Newton's second law:F=dtdp
For constant mass:
F=mdtdv=ma
Conservation of Momentum
When Fext=0:
dtdptotal=0
ptotal=constant
For a system of particles:
ptotal=∑imiviconstant
Impulse
J=∫t1t2Fdt=Δp
Variable Force Example
If F(t)=F0sin(ωt) from t=0 to t=π/ω:
J=∫0π/ωF0sin(ωt)dt=ωF0[−cos(ωt)]0π/ω
J=ωF0(1−(−1))=ω2F0
Elastic Collisions (1D)
Conservation of momentum:
m1v1i+m2v2i=m1v1f+m2v2f
Conservation of kinetic energy:
21m1v1i2+21m2v2i2=21m1v1f2+21m2v2f2
(This energy is converted to heat, deformation, sound, etc.)
Coefficient of Restitution
For partially elastic collisions:
e=∣v1i−v2i∣∣v2f−v1f∣
e=1: perfectly elastic
e=0: perfectly inelastic
0<e<1: partially elastic
Relative velocity after collision:v2f−v1f=−e(v2i−v1i)
2D Collisions
Momentum conserved in each direction:
x-component:m1v1ix+m2v2ix=m1v1fx+m2v2fx
y-component:m1v1iy+m2v2iy=m1v1fy+m2v2fy
For elastic collisions, also conserve kinetic energy.
Rocket Motion
Variable mass: m(t) decreases as fuel burns.
mdtdv=−vreldtdm+Fext
where vrel is exhaust speed relative to rocket.
In space (Fext=0):
mdv=−vreldm
∫v0vdv′=−vrel∫m0mm
Δv=vrellnmm0
Tsiolkovsky equation
m1v1i+m2v2i=(m1+m2)vf
(1500)(20)+0=(1500+1000)vf
30000=2500vf
vf=12 m/s
(b) Initial kinetic energy:
KEi=21m1v1i2+0=21(1500)(20)2
KEi=300,000 J=300 kJ
(c) Final kinetic energy:
KEf=21(m1+m2)vf2=21(2500)(12)2
KEf=180,000 J=180 kJ
(d) Energy lost:
ΔKE=KEi−KEf=300−180
ΔKE=120 kJ
This energy is converted to heat, sound, and deformation.
2Problem 2medium
❓ Question:
A 2.0 kg block moving at 5.0 m/s collides elastically with a 3.0 kg block at rest. Find: (a) the final velocities of both blocks, (b) verify momentum conservation, and (c) verify kinetic energy conservation.
💡 Show Solution
Given:
m₁ = 2.0 kg, v₁ᵢ = 5.0 m/s
m₂ = 3.0 kg, v₂ᵢ = 0
Elastic collision
(a) Final velocities:
For elastic collision in 1D:
v1f=m1+m2
v1f=5.0−1.0
(Block 1 bounces backward)
v2f=m
v2f=5.04.0
(b) Momentum conservation check:
Initial: pi=(2.0)(5.0)+0=10 kg·m/s
Final: pf=(2.0)(−1.0)+(3.0)(4.0)=−2 kg·m/s ✓
(c) Kinetic energy conservation check:
Initial:
KEi=21(2.0)(
Final:
KEf=21
KEf=1+24=25 J ✓
3Problem 3hard
❓ Question:
A ballistic pendulum consists of a block of mass M = 2.0 kg hanging from strings of length L = 1.5 m. A bullet of mass m = 0.01 kg is fired horizontally into the block at velocity v₀. After the collision, the block (with bullet embedded) swings up to a maximum angle θ = 40°. Find: (a) the velocity just after collision, (b) the initial bullet velocity, and (c) the percentage of energy lost.
💡 Show Solution
Given:
M = 2.0 kg
m = 0.01 kg
L = 1.5 m
θ = 40°
(a) Velocity just after collision:
After collision, block+bullet swing to height h:
h=L(1−cosθ)=1.5(1−cos40°)
Most energy goes into deformation, heat, and sound!
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Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.