Kinetics of Particles: Impulse and Momentum

Principle

Linear impulse of a force is $\int \mathbf{F}\,dt$ (unit N·s). Linear momentum is $m\mathbf{v}$ (unit kg·m/s). The principle states that the linear impulse of the resultant force acting on a particle during a time interval equals the change in its linear momentum:

$$m\mathbf{v}_1 + \int_{t_1}^{t_2} \mathbf{F}\,dt = m\mathbf{v}_2$$

Derivation

Newton's second law: $\mathbf{F} = m\dfrac{d\mathbf{v}}{dt}$ (constant mass). Multiply by $dt$ and integrate from $t_1$ to $t_2$:

$$\int_{t_1}^{t_2}\mathbf{F}\,dt = m\int_{\mathbf{v}_1}^{\mathbf{v}_2} d\mathbf{v} = m\mathbf{v}_2 - m\mathbf{v}_1$$

It is a vector equation, so it is applied separately in each direction. It is useful when forces depend on time and when time (not distance) is required.

Impulsive force

An impulsive force is a very large force acting for a very short time (for example in a collision or a hammer blow), so that its impulse is finite while the displacement during the action is negligible. Non-impulsive forces such as weight and spring forces are ignored during the impact.

Conservation of linear momentum

For a system of particles, the internal forces occur in equal and opposite pairs (Newton's third law) and their impulses cancel. If the external impulse is zero in a direction:

$$\sum m_i \mathbf{u}_i = \sum m_i \mathbf{v}_i$$

the total momentum of the system in that direction is the same before and after, for example in an explosion, a recoiling gun, or a collision between two bodies.

Direct central impact

Two bodies collide in direct central impact when their centres of mass lie on the line of impact, which is also the common normal at the contact point, and the velocities before and after impact are along this line. If the surfaces are smooth, the impulses act only along the line of impact.

  before:  (m1)--> u1      (m2)--> u2      u1 > u2
  after :  (m1)--> v1      (m2)--> v2      v2 >= v1

The impact has two phases: deformation (from first contact to maximum compression, when both bodies have a common velocity) and restitution (the bodies regain shape and separate).

Coefficient of restitution

$e$ is the ratio of the impulse during restitution to the impulse during deformation, which for a direct central impact equals

$$e = \frac{\text{velocity of separation}}{\text{velocity of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$$

$e = 1$ for a perfectly elastic impact and $e = 0$ for a perfectly plastic impact; $0 < e < 1$ in practice.

Final velocities

Conservation of momentum (no external impulse):

$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad (1)$$

Definition of $e$: $v_2 - v_1 = e(u_1 - u_2)$ (2)

Solving (1) and (2):

$$v_1 = \frac{m_1 u_1 + m_2 u_2 - m_2 e(u_1 - u_2)}{m_1 + m_2} \qquad v_2 = \frac{m_1 u_1 + m_2 u_2 + m_1 e(u_1 - u_2)}{m_1 + m_2}$$

Loss of kinetic energy

$$\Delta T = \tfrac12 m_1 u_1^2 + \tfrac12 m_2 u_2^2 - \tfrac12 m_1 v_1^2 - \tfrac12 m_2 v_2^2$$

Substituting the velocities and simplifying:

$$\Delta T = \frac{m_1 m_2}{2(m_1 + m_2)}\,(1 - e^2)\,(u_1 - u_2)^2$$

For $e = 1$, $\Delta T = 0$ (kinetic energy conserved); for $e = 0$ the loss is maximum and both bodies move together with velocity $(m_1u_1 + m_2u_2)/(m_1+m_2)$. The lost energy becomes heat, sound and permanent deformation.

Given: $m_A = 5$ kg, $u_A = 6$ m/s, $m_B = 3$ kg, $u_B = 2$ m/s, $e = 0.7$ (both to the right).

(a) Velocities after impact

Conservation of momentum: total momentum $= 5(6) + 3(2) = 36.0$ kg·m/s.

$$5v_A + 3v_B = 36 \quad (1)$$

Restitution: $v_B - v_A = e(u_A - u_B) = 0.7(6 - 2) = 2.8$ (2)

From (2), $v_B = v_A + 2.8$. Put in (1): $5v_A + 3v_A + 8.4 = 36$, so $8v_A = 27.6$.

$$v_A = 3.45\ \text{m/s} \qquad v_B = 6.25\ \text{m/s}$$

Both move to the right; B now moves faster than A, so there is no further impact.

(b) Loss of kinetic energy

$$\Delta T = 96.00 - 88.35 = 7.65\ \text{J}$$

Check by formula: $\dfrac{5 \times 3}{2 \times 8}(1 - 0.49)(4)^2 = 0.9375 \times 0.51 \times 16 = 7.65$ J.

(c) Impulse on B

Impulse equals change of momentum of B:

$$I = m_B(v_B - u_B) = 3(6.25 - 2) = 12.75\ \text{N·s}$$

(The impulse on A is equal and opposite, $5(3.45 - 6) = -12.75$ N·s.)

Answer: v_A = 3.45 m/s, v_B = 6.25 m/s; energy lost = 7.65 J; impulse on B = 12.75 N·s.

Given: $m = 0.4$ kg, $u = 20$ m/s at $30^\circ$ to the wall surface, $e = 0.75$. The wall is smooth, so the impulse acts along the normal only.

   wall |
        |  normal (n)
        | <--
   -----+-------> tangent (t)
        |  \  incoming 30 deg to wall

Components of initial velocity:

• Normal (perpendicular to wall): $u_n = 20\sin30^\circ = 10.00$ m/s
• Tangential (along wall): $u_t = 20\cos30^\circ = 17.32$ m/s

Velocity after impact

• Tangential component is unchanged (no tangential impulse): $v_t = 17.32$ m/s
• Normal component reverses with restitution: $v_n = e\,u_n = 0.75 \times 10 = 7.50$ m/s

$$v = \sqrt{v_n^2 + v_t^2} = \sqrt{7.5^2 + 17.32^2} = 18.87\ \text{m/s}$$

Direction with the wall: $\tan\beta = v_n/v_t = 7.5/17.32$, so $\beta = 23.4^\circ$ (the ball leaves at $23.4^\circ$ to the wall, a flatter angle than $30^\circ$).

Impulse and average force

Impulse on the ball along the normal: $I = m(v_n + u_n) = m(1 + e)u_n = 0.4 \times 1.75 \times 10 = 7.00$ N·s. The impulse on the wall is equal and opposite.

$$F_{avg} = \frac{I}{\Delta t} = \frac{7.00}{0.01} = 700\ \text{N}$$

Kinetic energy lost

$$\Delta T = \tfrac12(0.4)(20^2 - 356.25) = 8.75\ \text{J}$$

Answer: v = 18.87 m/s at 23.4° to the wall; impulse = 7.0 N·s; average force = 700 N; energy lost = 8.75 J.

Given: bullet $m = 0.020$ kg, $u = 400$ m/s; block $M = 2$ kg; string length $l = 1.5$ m.

         O
         |  l
   --->  |
  bullet [M]   -> swings up through h

(a) Velocity just after impact

The impact is very short, so the string tension and weight are non-impulsive. Linear momentum is conserved horizontally:

$$0.020 \times 400 = (0.020 + 2)\,v \;\Rightarrow\; v = \frac{8.0}{2.02} = 3.960\ \text{m/s}$$

(b) Height of rise

After impact, mechanical energy is conserved:

$$\tfrac12 v^2 = g h \;\Rightarrow\; h = \frac{(3.960)^2}{2 \times 9.81} = 0.799\ \text{m}$$

(c) Angle of swing

$$\cos\theta = \frac{l - h}{l} = \frac{1.5 - 0.799}{1.5} = 0.4670 \;\Rightarrow\; \theta = 62.2^\circ$$

(d) Kinetic energy lost

• Before: $\tfrac12(0.020)(400)^2 = 1600$ J
• After: $\tfrac12(2.02)(3.960)^2 = 15.84$ J
• Loss $= 1584.16$ J, which is $\dfrac{1584.16}{1600} \times 100 = 99.0\ \%$

The loss is due to the bullet penetrating and heating the wood (plastic impact, $e = 0$).

Answer: v = 3.96 m/s; h = 799 mm; θ = 62.2°; KE lost = 99.0 %.