Momentum and Collisions

Conservation of momentum, elastic and inelastic collisions

Momentum and Collisions

Linear Momentum

p⃗=mv⃗\vec{p} = m\vec{v}

Newton's second law: F⃗=dp⃗dt\vec{F} = \frac{d\vec{p}}{dt}

For constant mass: F⃗=mdv⃗dt=ma⃗\vec{F} = m\frac{d\vec{v}}{dt} = m\vec{a}

Conservation of Momentum

When F⃗ext=0\vec{F}_{ext} = 0:

dp⃗totaldt=0\frac{d\vec{p}_{total}}{dt} = 0

p⃗total=constant\vec{p}_{total} = \text{constant}

For a system of particles: p⃗total=∑imiv⃗i=constant\vec{p}_{total} = \sum_i m_i\vec{v}_i = \text{constant}

Impulse

J⃗=∫t1t2F⃗ dt=Δp⃗\vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt = \Delta \vec{p}

Variable Force Example

If F(t)=F0sin⁡(ωt)F(t) = F_0\sin(\omega t) from t=0t = 0 to t=π/ωt = \pi/\omega:

J=∫0π/ωF0sin⁡(ωt) dt=F0ω[−cos⁡(ωt)]0π/ωJ = \int_0^{\pi/\omega} F_0\sin(\omega t) \, dt = \frac{F_0}{\omega}[-\cos(\omega t)]_0^{\pi/\omega}

J=F0ω(1−(−1))=2F0ωJ = \frac{F_0}{\omega}(1 - (-1)) = \frac{2F_0}{\omega}

Elastic Collisions (1D)

Conservation of momentum: m1v1i+m2v2i=m1v1f+m2v2fm_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

Conservation of kinetic energy: 12m1v1i2+12m2v2i2=12m1v1f2+12m2v2f2\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2

Solutions: v1f=(m1−m2)v1i+2m2v2im1+m2v_{1f} = \frac{(m_1 - m_2)v_{1i} + 2m_2v_{2i}}{m_1 + m_2}

v2f=(m2−m1)v2i+2m1v1im1+m2v_{2f} = \frac{(m_2 - m_1)v_{2i} + 2m_1v_{1i}}{m_1 + m_2}

Special Cases

Equal masses (m1=m2m_1 = m_2): v1f=v2i,v2f=v1iv_{1f} = v_{2i}, \quad v_{2f} = v_{1i} (velocities exchange)

Target at rest (v2i=0v_{2i} = 0): v1f=m1−m2m1+m2v1i,v2f=2m1m1+m2v1iv_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i}, \quad v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i}

Massive target (m2≫m1m_2 \gg m_1, v2i=0v_{2i} = 0): v1f≈−v1i,v2f≈0v_{1f} \approx -v_{1i}, \quad v_{2f} \approx 0 (light object bounces back)

Inelastic Collisions

Momentum conserved, but kinetic energy not conserved.

Perfectly inelastic (objects stick together): m1v1i+m2v2i=(m1+m2)vfm_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f

vf=m1v1i+m2v2im1+m2v_f = \frac{m_1v_{1i} + m_2v_{2i}}{m_1 + m_2}

Energy lost: ΔKE=KEf−KEi=12m1m2m1+m2(v1i−v2i)2\Delta KE = KE_f - KE_i = \frac{1}{2}\frac{m_1m_2}{m_1 + m_2}(v_{1i} - v_{2i})^2

(This energy is converted to heat, deformation, sound, etc.)

Coefficient of Restitution

For partially elastic collisions: e=∣v2f−v1f∣∣v1i−v2i∣e = \frac{|v_{2f} - v_{1f}|}{|v_{1i} - v_{2i}|}

  • e=1e = 1: perfectly elastic
  • e=0e = 0: perfectly inelastic
  • 0<e<10 < e < 1: partially elastic

Relative velocity after collision: v2f−v1f=−e(v2i−v1i)v_{2f} - v_{1f} = -e(v_{2i} - v_{1i})

2D Collisions

Momentum conserved in each direction:

x-component: m1v1ix+m2v2ix=m1v1fx+m2v2fxm_1v_{1ix} + m_2v_{2ix} = m_1v_{1fx} + m_2v_{2fx}

y-component: m1v1iy+m2v2iy=m1v1fy+m2v2fym_1v_{1iy} + m_2v_{2iy} = m_1v_{1fy} + m_2v_{2fy}

For elastic collisions, also conserve kinetic energy.

Rocket Motion

Variable mass: m(t)m(t) decreases as fuel burns.

mdvdt=−vreldmdt+Fextm\frac{dv}{dt} = -v_{rel}\frac{dm}{dt} + F_{ext}

where vrelv_{rel} is exhaust speed relative to rocket.

In space (Fext=0F_{ext} = 0): m dv=−vrel dmm \, dv = -v_{rel} \, dm

∫v0vdv′=−vrel∫m0mdm′m′\int_{v_0}^v dv' = -v_{rel}\int_{m_0}^m \frac{dm'}{m'}

Δv=vrelln⁡m0m\Delta v = v_{rel}\ln\frac{m_0}{m}

Tsiolkovsky equation

📚 Practice Problems

1Problem 1easy

❓ Question:

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.

💡 Show Solution

Given:

  • m₁ = 1500 kg, v₁ᵢ = 20 m/s
  • m₂ = 1000 kg, v₂ᵢ = 0
  • Perfectly inelastic collision

(a) Final velocity:

Conservation of momentum: m1v1i+m2v2i=(m1+m2)vfm_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)v_f

(1500)(20)+0=(1500+1000)vf(1500)(20) + 0 = (1500 + 1000)v_f

30000=2500vf30000 = 2500v_f

vf=12 m/s\boxed{v_f = 12 \text{ m/s}}

(b) Initial kinetic energy:

KEi=12m1v1i2+0=12(1500)(20)2KE_i = \frac{1}{2}m_1 v_{1i}^2 + 0 = \frac{1}{2}(1500)(20)^2

KEi=300,000 J=300 kJ\boxed{KE_i = 300,000 \text{ J} = 300 \text{ kJ}}

(c) Final kinetic energy:

KEf=12(m1+m2)vf2=12(2500)(12)2KE_f = \frac{1}{2}(m_1 + m_2)v_f^2 = \frac{1}{2}(2500)(12)^2

KEf=180,000 J=180 kJ\boxed{KE_f = 180,000 \text{ J} = 180 \text{ kJ}}

(d) Energy lost:

ΔKE=KEi−KEf=300−180\Delta KE = KE_i - KE_f = 300 - 180

ΔKE=120 kJ\boxed{\Delta KE = 120 \text{ kJ}}

This energy is converted to heat, sound, and deformation.

2Problem 2easy

❓ Question:

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.

💡 Show Solution

Given:

  • m₁ = 1500 kg, v₁ᵢ = 20 m/s
  • m₂ = 1000 kg, v₂ᵢ = 0
  • Perfectly inelastic collision

(a) Final velocity:

Conservation of momentum: m1v1i+m2v2i=(m1+m2)vfm_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)v_f

(1500)(20)+0=(1500+1000)vf(1500)(20) + 0 = (1500 + 1000)v_f

30000=2500vf30000 = 2500v_f

vf=12 m/s\boxed{v_f = 12 \text{ m/s}}

(b) Initial kinetic energy:

KEi=12m1v1i2+0=12(1500)(20)2KE_i = \frac{1}{2}m_1 v_{1i}^2 + 0 = \frac{1}{2}(1500)(20)^2

KEi=300,000 J=300 kJ\boxed{KE_i = 300,000 \text{ J} = 300 \text{ kJ}}

(c) Final kinetic energy:

KEf=12(m1+m2)vf2=12(2500)(12)2KE_f = \frac{1}{2}(m_1 + m_2)v_f^2 = \frac{1}{2}(2500)(12)^2

KEf=180,000 J=180 kJ\boxed{KE_f = 180,000 \text{ J} = 180 \text{ kJ}}

(d) Energy lost:

ΔKE=KEi−KEf=300−180\Delta KE = KE_i - KE_f = 300 - 180

ΔKE=120 kJ\boxed{\Delta KE = 120 \text{ kJ}}

This energy is converted to heat, sound, and deformation.

3Problem 3medium

❓ 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−m2m1+m2v1i=2.0−3.02.0+3.0(5.0)v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i} = \frac{2.0 - 3.0}{2.0 + 3.0}(5.0)

v1f=−1.05.0(5.0)=−1.0 m/sv_{1f} = \frac{-1.0}{5.0}(5.0) = \boxed{-1.0 \text{ m/s}}

(Block 1 bounces backward)

v2f=2m1m1+m2v1i=2(2.0)5.0(5.0)v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i} = \frac{2(2.0)}{5.0}(5.0)

v2f=4.05.0(5.0)=4.0 m/sv_{2f} = \frac{4.0}{5.0}(5.0) = \boxed{4.0 \text{ m/s}}

(b) Momentum conservation check:

Initial: pi=(2.0)(5.0)+0=10p_i = (2.0)(5.0) + 0 = 10 kg·m/s

Final: pf=(2.0)(−1.0)+(3.0)(4.0)=−2+12=10p_f = (2.0)(-1.0) + (3.0)(4.0) = -2 + 12 = 10 kg·m/s ✓

(c) Kinetic energy conservation check:

Initial: KEi=12(2.0)(5.0)2=25 JKE_i = \frac{1}{2}(2.0)(5.0)^2 = 25 \text{ J}

Final: KEf=12(2.0)(−1.0)2+12(3.0)(4.0)2KE_f = \frac{1}{2}(2.0)(-1.0)^2 + \frac{1}{2}(3.0)(4.0)^2

KEf=1+24=25 JKE_f = 1 + 24 = 25 \text{ J} ✓

4Problem 4medium

❓ 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−m2m1+m2v1i=2.0−3.02.0+3.0(5.0)v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i} = \frac{2.0 - 3.0}{2.0 + 3.0}(5.0)

v1f=−1.05.0(5.0)=−1.0 m/sv_{1f} = \frac{-1.0}{5.0}(5.0) = \boxed{-1.0 \text{ m/s}}

(Block 1 bounces backward)

v2f=2m1m1+m2v1i=2(2.0)5.0(5.0)v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i} = \frac{2(2.0)}{5.0}(5.0)

v2f=4.05.0(5.0)=4.0 m/sv_{2f} = \frac{4.0}{5.0}(5.0) = \boxed{4.0 \text{ m/s}}

(b) Momentum conservation check:

Initial: pi=(2.0)(5.0)+0=10p_i = (2.0)(5.0) + 0 = 10 kg·m/s

Final: pf=(2.0)(−1.0)+(3.0)(4.0)=−2+12=10p_f = (2.0)(-1.0) + (3.0)(4.0) = -2 + 12 = 10 kg·m/s ✓

(c) Kinetic energy conservation check:

Initial: KEi=12(2.0)(5.0)2=25 JKE_i = \frac{1}{2}(2.0)(5.0)^2 = 25 \text{ J}

Final: KEf=12(2.0)(−1.0)2+12(3.0)(4.0)2KE_f = \frac{1}{2}(2.0)(-1.0)^2 + \frac{1}{2}(3.0)(4.0)^2

KEf=1+24=25 JKE_f = 1 + 24 = 25 \text{ J} ✓

5Problem 5hard

❓ 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−cos⁡θ)=1.5(1−cos⁡40°)h = L(1 - \cos\theta) = 1.5(1 - \cos 40°)

h=1.5(1−0.766)=1.5(0.234)=0.351 mh = 1.5(1 - 0.766) = 1.5(0.234) = 0.351 \text{ m}

Energy conservation (after collision): 12(M+m)v12=(M+m)gh\frac{1}{2}(M + m)v_1^2 = (M + m)gh

v1=2gh=2(9.8)(0.351)v_1 = \sqrt{2gh} = \sqrt{2(9.8)(0.351)}

v1=2.62 m/s\boxed{v_1 = 2.62 \text{ m/s}}

(b) Initial bullet velocity:

Momentum conservation (during collision): mv0=(M+m)v1mv_0 = (M + m)v_1

v0=(M+m)v1m=(2.01)(2.62)0.01v_0 = \frac{(M + m)v_1}{m} = \frac{(2.01)(2.62)}{0.01}

v0=527 m/s\boxed{v_0 = 527 \text{ m/s}}

(c) Energy lost percentage:

Initial KE: KEi=12mv02=12(0.01)(527)2=1389 JKE_i = \frac{1}{2}mv_0^2 = \frac{1}{2}(0.01)(527)^2 = 1389 \text{ J}

KE just after collision: KEf=12(M+m)v12=12(2.01)(2.62)2=6.90 JKE_f = \frac{1}{2}(M + m)v_1^2 = \frac{1}{2}(2.01)(2.62)^2 = 6.90 \text{ J}

Percentage lost: KEi−KEfKEi×100%=1389−6.901389×100%\frac{KE_i - KE_f}{KE_i} \times 100\% = \frac{1389 - 6.90}{1389} \times 100\%

99.5% of energy lost\boxed{99.5\% \text{ of energy lost}}

Most energy goes into deformation, heat, and sound!

6Problem 6hard

❓ 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−cos⁡θ)=1.5(1−cos⁡40°)h = L(1 - \cos\theta) = 1.5(1 - \cos 40°)

h=1.5(1−0.766)=1.5(0.234)=0.351 mh = 1.5(1 - 0.766) = 1.5(0.234) = 0.351 \text{ m}

Energy conservation (after collision): 12(M+m)v12=(M+m)gh\frac{1}{2}(M + m)v_1^2 = (M + m)gh

v1=2gh=2(9.8)(0.351)v_1 = \sqrt{2gh} = \sqrt{2(9.8)(0.351)}

v1=2.62 m/s\boxed{v_1 = 2.62 \text{ m/s}}

(b) Initial bullet velocity:

Momentum conservation (during collision): mv0=(M+m)v1mv_0 = (M + m)v_1

v0=(M+m)v1m=(2.01)(2.62)0.01v_0 = \frac{(M + m)v_1}{m} = \frac{(2.01)(2.62)}{0.01}

v0=527 m/s\boxed{v_0 = 527 \text{ m/s}}

(c) Energy lost percentage:

Initial KE: KEi=12mv02=12(0.01)(527)2=1389 JKE_i = \frac{1}{2}mv_0^2 = \frac{1}{2}(0.01)(527)^2 = 1389 \text{ J}

KE just after collision: KEf=12(M+m)v12=12(2.01)(2.62)2=6.90 JKE_f = \frac{1}{2}(M + m)v_1^2 = \frac{1}{2}(2.01)(2.62)^2 = 6.90 \text{ J}

Percentage lost: KEi−KEfKEi×100%=1389−6.901389×100%\frac{KE_i - KE_f}{KE_i} \times 100\% = \frac{1389 - 6.90}{1389} \times 100\%

99.5% of energy lost\boxed{99.5\% \text{ of energy lost}}

Most energy goes into deformation, heat, and sound!

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