Linear momentum and impulse

Newton stated his second law for the linear momentum \(\bm G = m\bm v\), the “quantity of motion” of the epigraph,

\[ \sum\bm F = \frac{d\bm G}{dt} , \]

which for a constant mass is \(\sum\bm F = m\bm a\). Integrating from \(t_1\) to \(t_2\) gives

\[ \bm G_1 + \int_{t_1}^{t_2}\sum\bm F\,dt = \bm G_2 , \qquad \text{i.e.}\qquad m\bm v_1 + \int_{t_1}^{t_2}\sum\bm F\,dt = m\bm v_2 . \tag{7.5.1}\]

The time integral of a force is its impulse, measured in newton seconds. 7.5.1 is a vector equation, one scalar equation per direction. When only the change of velocity and the duration \(\Delta t\) are known, it gives the average force \(\bar{\bm F} = \Delta\bm G/\Delta t\).

The forces of an impact are impulsive: very large and acting for a very short time. Their impulse is finite, of the order of the momentum change. The weight, a spring or a soft support stay of ordinary size, and their impulse during the impact is that ordinary force times \(\Delta t\), which vanishes as \(\Delta t\) does. Over the duration of an impact these non-impulsive forces are left out, and checking that they are small is part of every impact problem.

Example 1: Clearing the ball off the goal line

A football of mass \(m = 0.43\) kg rolls towards the goal line at \(v_1 = 15\) m/s. A defender kicks it away at \(v_2 = 25\) m/s, at the angle \(\theta = 30^\circ\) to the ground (Figure 7.5.1). The foot is in contact with the ball for \(\Delta t = 10\) ms. Find the average force of the kick.

With \(x\) along the ground away from the goal and \(y\) up, the velocities are \(\bm v_1 = [-v_1,\ 0]^\mathsf{T}\) and \(\bm v_2 = v_2[\cos\theta,\ \sin\theta]^\mathsf{T}\). Leaving out the weight for now, 7.5.1 gives the impulse of the kick and its average force,

\[ \bm J = \int_0^{\Delta t}\bm F\,dt = m\bm v_2 - m\bm v_1 , \qquad \bar{\bm F} = \frac{\bm J}{\Delta t} . \]

m, v1, v2, th, dt = 0.43, 15.0, 25.0, np.radians(30), 0.010
J = m*v2*np.array([np.cos(th), np.sin(th)]) - m*np.array([-v1, 0.0])
F_avg = J/dt
J_weight = m*9.81*dt

With the numbers of the example,

The foot pushes on the ball with 1.67 kN on average, about 400 times the weight of the ball, and mostly along the ground, since most of the momentum change is the reversal of \(v_1\). The impulse of the weight over the 10 ms, 0.04 N·s, is a quarter of a percent of the impulse of the kick, which justifies leaving it out. The force is not constant during the contact: it rises from zero as the ball deforms and falls back to zero as it leaves. For a pulse shaped like half a sine wave the peak is \(\pi/2\) times the average, about 2.6 kN.

Conservation of momentum

For a system of bodies the forces between them come in equal and opposite pairs, and so do their impulses. Summing 7.5.1 over the bodies, the internal impulses cancel, and when the external forces have no impulse, or a negligible one,

\[ \sum_i m_i\bm v_{i1} = \sum_i m_i\bm v_{i2} . \tag{7.5.2}\]

This is the epigraph: the total momentum “suffers no change from the action of bodies among themselves”. In a collision the forces between the bodies are impulsive and the external ones usually are not, so the momentum of the colliding bodies is the same just before and just after the impact, whatever happens during it.

Central impact

Two bodies moving along one line collide (Figure 7.5.2 shows one case). During the contact they first press into each other until they move with the same velocity, the deformation phase, and then push apart, the restitution phase. Conservation of momentum along the line gives

\[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' , \]

one equation for the two velocities \(v_1'\) and \(v_2'\) after the impact. The second equation comes from the materials. How much of the deformation is given back is measured by the coefficient of restitution, the ratio of the speed of separation after the impact to the speed of approach before it,

\[ e = \frac{v_2' - v_1'}{v_1 - v_2} , \qquad 0 \leq e \leq 1 . \tag{7.5.3}\]

With \(e = 1\) the impact is perfectly elastic and no kinetic energy is lost; with \(e = 0\) it is perfectly plastic and the bodies move on together. The coefficient depends on the materials, on their shape and on the impact speed, and it is measured, not derived.

m1, m2, u1, u2, e = sp.symbols('m_1 m_2 v_1 v_2 e', real=True)
w1, w2 = sp.symbols("v_1' v_2'", real=True)
sol_c = sp.solve([sp.Eq(m1*u1 + m2*u2, m1*w1 + m2*w2), sp.Eq(w2 - w1, e*(u1 - u2))], [w1, w2], dict=True)[0]
dT = sp.factor(sp.simplify((m1*u1**2 + m2*u2**2 - m1*sol_c[w1]**2 - m2*sol_c[w2]**2)/2))

Solving the two equations, the velocities after the impact and the kinetic energy lost in it are

The energy lost grows with the square of the approach speed and with \(1 - e^2\), and it vanishes only for \(e = 1\).

Example 2: A ball dropped on a block

A steel ball of mass \(m_1 = 0.1\) kg is dropped from the height \(H = 1.0\) m onto a block of mass \(m_2 = 0.5\) kg that rests on a soft foam pad (Figure 7.5.2). The ball rebounds to the height \(h = 0.36\) m. Find the velocity of the block just after the impact, the coefficient of restitution, and the share of the kinetic energy lost.

g = 9.81
m1v, m2v, H, h = 0.1, 0.5, 1.0, 0.36
va = np.sqrt(2*g*H)
va_p = -np.sqrt(2*g*h)
vb_p = m1v*(va - va_p)/m2v
e_drop = (vb_p - va_p)/va
loss = 1 - (m1v*va_p**2 + m2v*vb_p**2)/(m1v*va**2)

Oblique impact

When the velocities are not along the line through the contact, we split them into the component along the common normal \(\bm e_n\) of the contact and the component along the tangent \(\bm e_t\) (Figure 7.5.3). If the surfaces are smooth, the contact force acts along \(\bm e_n\) only, which gives four conditions for two bodies:

1. the tangential velocity of each body is unchanged, \(v_{1t}' = v_{1t}\) and \(v_{2t}' = v_{2t}\);
2. the momentum along \(\bm e_n\) is conserved, \(m_1v_{1n} + m_2v_{2n} = m_1v_{1n}' + m_2v_{2n}'\);
3. the normal components obey the coefficient of restitution, 7.5.3 with \(n\) components.

For a ball bouncing off a fixed floor the momentum condition drops out, the floor being immovable, and what remains is \(v_t' = v_t\) and \(v_n' = -e\,v_n\). With the angles measured from the floor, \(\tan\theta' = e\tan\theta\): the ball leaves flatter than it arrived, and slower.

Example 3: A tennis ball bouncing across the floor

The rules of tennis fix the coefficient of restitution of the ball. A ball dropped from 254 cm onto concrete must rebound to more than 135 cm and less than 147 cm (ITF Rules of Tennis, Appendix I). Since the speed at impact is \(\sqrt{2gH}\) and the speed at rebound \(\sqrt{2gh}\), the drop test measures

\[ e = \frac{\sqrt{2gh}}{\sqrt{2gH}} = \sqrt{\frac{h}{H}} , \]

so an approved ball has \(0.728 < e < 0.761\). We take \(e = 0.745\). The ball rolls off a table of height \(h_0 = 0.9\) m with the horizontal speed \(u = 3\) m/s and bounces along a smooth floor. Find its path, the heights of the bounces, and when it stops bouncing.

def bounce(e=0.745, h0=0.9, u=3.0, g=9.81, T=3.0, dt=1e-5):
    n = int(T/dt)
    x, y, vy = np.zeros(n), np.zeros(n), np.zeros(n)
    y[0] = h0
    for i in range(n - 1):
        vy[i + 1] = vy[i] - g*dt
        y[i + 1] = y[i] + dt*vy[i + 1]
        x[i + 1] = x[i] + dt*u
        if y[i + 1] < 0 and vy[i + 1] < 0:         # the floor: reflect the normal velocity
            y[i + 1], vy[i + 1] = -y[i + 1], -e*vy[i + 1]
    return x, y, vy

e_t, h0, u0 = 0.745, 0.9, 3.0
xb, yb, vyb = bounce(e_t, h0, u0)
t0 = np.sqrt(2*h0/9.81)
t_inf = t0*(1 + e_t)/(1 - e_t)
peaks = np.where((vyb[:-1] > 0) & (vyb[1:] <= 0))[0]
h_sim = yb[peaks][:5]
h_pred = h0*e_t**(2*np.arange(1, 6))

fig, ax = plt.subplots(figsize=(6.5, 2.8))
ax.plot(xb, yb, color='C2', lw=1.4)
ax.plot(xb[peaks][:5], h_pred, 'ko', ms=4, label=r'$h_n = e^{2n}h_0$')
ax.axvline(u0*t_inf, color='0.5', ls='--', lw=1)
ax.text(u0*t_inf - 0.1, 0.35, r'$u\,t_\infty$', ha='right')
ax.set_xlabel('$x$ [m]'); ax.set_ylabel('$y$ [m]'); ax.grid(alpha=0.3)
ax.set_xlim(0, 9); ax.set_ylim(0, 0.9); ax.legend(loc='upper center', fontsize=9)
end_ticks(fig)
plt.tight_layout(); plt.show()

Example 4: The bounce as a contact model

The coefficient of restitution describes the impact by its result. A simulation program has to compute what happens during the contact, and the usual model is the one of Figure 7.5.4. While the ball presses into the floor by the penetration \(\delta\), a spring \(k\) and a damper \(c\) push it back; when they are apart, nothing acts. The contact can push but not pull, so the force is

\[ F = \max\bigl(0,\ k\delta + c\dot\delta\bigr) \quad\text{for } \delta > 0 , \qquad F = 0 \quad\text{otherwise} , \]

the force that switches on in the bungee jump of Particle kinetics, with a damper added.

We take the tennis ball of Example 3, of mass \(m = 57\) g and radius 33 mm, and a contact stiffness \(k = 3\cdot 10^4\) N/m, which gives a contact time of a few milliseconds. The damping is then chosen so that the model passes the ITF drop test: dropped from 254 cm, it must come back with \(e = 0.745\). With the height \(y\) of the centre and the penetration \(\delta = R - y\), the equation of motion during the drop is

\[ m\ddot y = -mg + F(\delta, \dot\delta) . \]

We find \(c\) by repeating the drop and adjusting \(c\) until the computed \(e\) is 0.745. Linear vibration theory gives a closed form for a spring and damper that also pull, \(e = \exp\bigl(-\zeta\pi/\sqrt{1 - \zeta^2}\bigr)\) with the damping ratio \(\zeta = c/(2\sqrt{km})\), which we compare with the calibrated model. The energy check is the work-energy theorem over the contact: the change of the ball’s kinetic energy between touching down and lifting off equals the work \(\int F\,\dot y\,dt\) of the contact force, since the weight does no net work between two instants at the same height.

from scipy.optimize import brentq

def drop_test(c, m=0.057, k=3e4, R=0.033, H=2.54, g=9.81, dt=2e-6):
    y, v, t = H + R, 0.0, 0.0
    hist, inside = [], False
    while True:
        delta = R - y
        F = max(k*delta - c*v, 0.0) if delta > 0 else 0.0     # pushes, never pulls
        if delta > 0 and not inside:
            inside, v_in = True, v
        if inside:
            hist.append((t, delta, F, v))
            if delta <= 0:
                h_arr = np.array(hist)
                return -v/v_in, h_arr
        v += dt*(F/m - g)
        y += dt*v
        t += dt

c_cal = brentq(lambda c: drop_test(c)[0] - 0.745, 0.1, 80)
e_cal, hist = drop_test(c_cal)
t_c, d_c, F_c, v_c = hist.T
zeta = c_cal/(2*np.sqrt(3e4*0.057))
e_linear = np.exp(-zeta*np.pi/np.sqrt(1 - zeta**2))
dT_contact = 0.057*(v_c[-1]**2 - v_c[0]**2)/2
W_contact = np.trapezoid(F_c*v_c, t_c)

The calibrated damper, the contact it produces, and the energy check are

fig, ax1 = plt.subplots(figsize=(6, 3))
ax1.plot(1e3*(t_c - t_c[0]), F_c, color='C3', lw=2, label='contact force $F$')
ax1.set_xlabel('$t$ [ms]'); ax1.set_ylabel('$F$ [N]'); ax1.grid(alpha=0.3)
ax1.set_xlim(0, 1e3*(t_c[-1] - t_c[0])); ax1.set_ylim(0, 280)
ax2 = ax1.twinx()
ax2.plot(1e3*(t_c - t_c[0]), 1e3*d_c, color='C0', lw=1.5, ls='--', label=r'penetration $\delta$')
ax2.set_ylabel(r'$\delta$ [mm]'); ax2.set_ylim(0, 10)
fig.legend(loc='upper right', bbox_to_anchor=(0.88, 0.93), fontsize=9)
end_ticks(fig)
plt.tight_layout(); plt.show()

A damping of about ten percent of critical reproduces the tennis ball: contact for 4.4 ms, a peak force of 258 N, more than 450 times the weight, and a rebound to 141 cm, inside the ITF window. The force peaks before the penetration does, because the damper adds to the spring while the ball is still moving in, and it reaches zero while the ball is still slightly pressed in: from then on the spring and the damper would have to pull, which a contact cannot do. That cut-off is why the calibrated model gives \(e = 0.745\) where the linear formula, which lets the damper pull, predicts 0.731. The force also jumps to about 58 N at the instant of touchdown, the damper’s response \(c\dot\delta\) to the impact speed, which a real ball does not show; contact models with a damping term proportional to the penetration, such as the Hunt-Crossley model, remove the jump. The kinetic energy lost in the contact, 0.63 J, equals the work of the contact force to within the accuracy of the time steps.

Angular impulse and momentum

The rotational counterpart follows from 7.6.2 of Rigid body kinetics, \(\sum M_G = \dot H_G\) with \(H_G = \bar I\omega\). Integrated over time,

\[ \bar I\omega_1 + \int_{t_1}^{t_2}\sum M_G\,dt = \bar I\omega_2 , \]

and the same holds about a fixed pin \(O\) with \(I_O\). The time integral of a moment is its angular impulse. Without an external moment the angular momentum is conserved. A figure skater who pulls in their arms reduces \(\bar I\), and since \(\bar I\omega\) stays the same the skater spins faster.

An impact driver uses angular impulse. Its motor does not turn the bit directly: it spins up a hammer, which strikes an anvil on the bit shaft many times per second. Each blow transfers the hammer’s angular momentum in a fraction of a millisecond, so the moment on the screw during the blow is many times the moment the motor could deliver steadily, while the moment the user’s hand has to hold back is only the small steady moment that spins up the hammer. The energy of each blow is small, but delivered in so short a time it comes out as a large moment, which loosens a screw that a drill of the same power only twists the user’s wrist against.

Further reading

Meriam, Kraige and Bolton treat impulse, momentum and impact for particles in chapter 3 and for rigid bodies in chapter 6 of their Dynamics [1].