Momentum and Impulse Notes (Chapter 2)

Momentum and Impulse — Chapter 2 Notes

  • Momentum (lowercase p) is defined as the product of mass and velocity: p = m\,v. The uppercase letter P is used for power, not momentum. Momentum is a vector quantity, built from mass and velocity.
    • Velocity is a vector; speed is a scalar. A velocity has magnitude, direction, and units. Example: |v| = 50\ \text{miles/hour} in the direction north; or |v| = 10\ \text{m/s} upward.
    • Because momentum is a vector, both magnitude and direction matter in calculations.
    • If an object isn’t moving, its momentum is zero: p = m v = m \cdot 0 = 0.
  • Why momentum matters in everyday life: momentum depends on both mass and velocity; heavier objects or faster objects have larger momentum and are harder to stop or change direction.
  • Impulse is the change in momentum and is related to force applied over a time interval.

Impulse

  • Impulse can be described in terms of average force over a contact time: \mathbf{J} = \mathbf{F}{\text{avg}}\,\Delta t. It is also the change in momentum: \mathbf{J} = \Delta \mathbf{p} = \mathbf{pf} - \mathbf{p_i}.
  • If two objects interact (a “hit” or collision), there is a force during a brief contact time. The impulse changes the momentum of the objects involved.
  • Real-world impulse often involves non-constant forces due to deformation during contact (e.g., ball hitting a bat, or players colliding).
    • The force-time profile frequently looks like it rises to a peak at maximum deformation and then falls as separation occurs.
    • Because the force is not constant, the impulse is best understood as the area under the force-vs-time curve: \text{Impulse} = \int{ti}^{tf} F(t)\, dt. For a simplifying assumption of constant force, this becomes \mathbf{J} = \mathbf{F}{\text{avg}}\,\Delta t.
  • Everyday impulse examples:
    • A linebacker collision where force and contact time determine the injury risk and momentum transfer.
    • A whiplash incident at a stoplight when a rear-end collision applies a large impulse over a very short time.
    • A baseball bat hitting a ball: the impulse from the bat on the ball changes the ball’s momentum; deformation causes the peak force.
  • Connection to momentum: the impulse delivered to one object is transferred to the other, causing a change in momentum (the momentum bookkeeping of the system).

Sign convention and vector changes

  • Momentum and impulse are vector quantities; signs matter.
  • A common pitfall is treating velocity changes as simple scalars. You must account for direction:
    • If a ball comes in with velocity (vi = -25\ \text{mph}) (negative meaning one direction) and leaves with (vf = +40\ \text{mph}), the change in velocity is not (vf - vi = 40 - 25 = 15) but rather
    • (\Delta v = vf - vi = 40 - (-25) = 65) mph, because you must use the signed velocities.
  • Example: If right is positive and a bat-ball collision changes velocity from (-25\,\text{mph}) to (+40\,\text{mph}), then
    • (\Delta v = 40 - (-25) = 65\ \text{mph}).
  • The momentum change is then (\Delta p = m \Delta v); direction follows the sign of the velocity change.
  • In one dimension, impulse and momentum changes obey: \mathbf{J} = \Delta \mathbf{p} = m(\mathbf{vf} - \mathbf{vi}).

Law of conservation of momentum

  • Momentum conservation: the total momentum of a system remains constant if no external impulse acts on the system. In mathematics: \sumi \mathbf{p}{i,\text{before}} = \sumi \mathbf{p}{i,\text{after}}.
  • System vs surroundings: the system consists of the objects of interest (e.g., baseball + bat); everything outside is surroundings. Momentum is conserved for the system when external forces are absent or net external impulse is zero.
  • Typically, we analyze two-body collisions; three-body collisions are rare in basic problems.
  • If the initial total momentum is zero, the final total momentum must also be zero:
    • Example: rifle and bullet before firing have zero momentum (both at rest). After firing, the bullet gains forward momentum and the rifle recoils with equal and opposite momentum to conserve total momentum.

Rifle and bullet example (conservation of momentum in a two-body process)

  • Given: rifle mass (mr = 1\,\text{kg}); bullet mass (mb = 0.005\,\text{kg}) (5 g).
  • Initial momenta: before firing, both are at rest, so
    • p{r,i} = mr v{r,i} = 1\cdot 0 = 0, p{b,i} = mb v{b,i} = 0.005\cdot 0 = 0. Total initial momentum: 0.
  • After firing, bullet velocity (v{b,f} = 300\ \text{m/s}). Bullet momentum: p{b,f} = mb v{b,f} = 0.005 \times 300 = 1.5\ \text{kg m/s}.
  • Recoil momentum of rifle must balance bullet momentum: p{r,f} = -p{b,f} = -1.5\ \text{kg m/s}. Therefore, rifle final velocity:
    • v{r,f} = \frac{p{r,f}}{m_r} = \frac{-1.5}{1} = -1.5\ \text{m/s}. (Negative sign indicates opposite direction to the bullet.)
  • Summary: momentum transfers from rifle to bullet via impulse, with equal and opposite momenta ensuring total momentum stays zero in the absence of external forces.

Baseball-bat contact and deformation (impulse in real contact)

  • In a baseball-bat collision, the force is not constant during contact; the ball and bat deform, producing a force-time profile that peaks when deformation is greatest.
  • The impulse received by the ball is the area under the force-time curve; a larger average force over a shorter time can produce a large impulse.
  • The ball’s momentum change is from its initial velocity to its final velocity after leaving the bat, incorporating the vector nature of velocity.
  • Visual intuition: if you cut and reassemble the impulse-time area into a rectangle of height equal to the average force and width equal to the contact time, you preserve the same impulse (area under the curve).
  • Broad takeaway: impulse captures the effect of a short, strong force during a collision and explains why forces during sports collisions are brief but powerful.

Systems: external forces and rotational motion

  • External forces: if external forces act during the interaction, momentum may not be conserved for the chosen system.
  • Rotational analog: conservation of angular momentum is the rotational counterpart to linear momentum conservation.
    • Angular momentum is (L = I \omega), where (I) is the moment of inertia and (\omega) is angular velocity.
    • If no external torque acts, total angular momentum is conserved: \sumi Li^{\text{before}} = \sumi Li^{\text{after}}.
    • Skater example: when an ice skater pulls arms in, their moment of inertia decreases and their angular speed increases to conserve angular momentum ((I\omega) stays constant). A diver pulling into a tuck behaves similarly.
  • This mirrors the linear momentum idea: mass distribution and velocity (or angular velocity) adjust to conserve the total momentum-like quantity when external interactions are internal to the system.

Quick recap of key ideas and connections

  • Momentum: p = m v; vector quantity with magnitude and direction.
  • Impulse: change in momentum; \mathbf{J} = \Delta \mathbf{p} = \mathbf{pf} - \mathbf{pi}, and for constant force, \mathbf{J} = \mathbf{F}{\text{avg}} \Delta t. For general cases, \mathbf{J} = \int{ti}^{tf} \mathbf{F}(t)\, dt.
  • Conservation of momentum: in a closed system with no external impulse, total momentum is constant: \sum \mathbf{p}{\text{before}} = \sum \mathbf{p}{\text{after}}.
  • Real-world collisions involve deformation and non-constant forces, but the impulse-momentum framework still applies via area under the force-time curve.
  • Sign conventions matter: always track direction; velocity changes must respect vector signs (e.g., a 25 mph approach and a 40 mph rebound are not simply added—one direction is negative, the other positive).
  • Two-body focus: most basic momentum problems involve two bodies; three-body problems are less common in introductory material.
  • Rotational momentum extends the same logic to spinning objects: angular momentum is conserved when external torques are absent; changing the distribution of mass changes the rotational speed to keep L constant.

Quick mental checklist for solving momentum problems

  • Identify the system (objects of interest) and the surroundings.
  • Determine whether external forces/torques are present or negligible.
  • Write the momentum before for all system components and the momentum after.
  • Use vector signs carefully for each component of velocity or angular velocity.
  • Apply Math: for linear momentum, solve for unknown final velocities from \sum \mathbf{p}{\text{before}} = \sum \mathbf{p}{\text{after}}. If needed, use the impulse relation \mathbf{J} = \Delta \mathbf{p} = \mathbf{F}{\text{avg}}\Delta t. If force is constant, this reduces to \mathbf{J} = \mathbf{F}{\text{avg}}\Delta t. In the general case, use \mathbf{J} = \int F(t)\, dt.
  • For rotational cases, apply L = I \omega and conserve L when external torque is zero.

Looking ahead

  • Next class will cover more applications, a short quiz, and additional momentum problems.