Exercises, examples & solutions - Classical Mechanics & Special Relativity for Starters

7.6.1Examples¶

7.6.1.1Newton’s Cradle¶

Click on the image below for a video on Newton’s cradle (gives you also the possibility to ‘play’ with different numerical solvers, from (too) simple to advanced).

7.6.1.2Elastic Collision¶

7.6.1.2.11D elastic collision¶

Consider two particles, $m_1$ and $m_2$ , moving along the $x$ -axis. The two particles will elastically collide. We set mass 2 at a value of 1 (kg) and vary $m_1$ . In the widget below, you can change the value of $m_1$ and of the velocities of both particles before the collision.

Solve the collision by using conservation of momentum and kinetic energy and compare your results with those of the widget.

import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
from math import atan2, degrees
from IPython.display import HTML

# -------------------------------
# Adjustable Parameters
# -------------------------------
m1 = 1   # mass of particle 1
m2 = 6   # mass of particle 2
v1x = 1  # x velocity of particle 1
v1y = 0  # y velocity of particle 1
v2y = 1  # y velocity of particle 2

# -------------------------------
# Constants and Initial Velocities
# -------------------------------
dt = 0.05
t_stop = 10
tcoll = 5
scale = 40

v1 = np.array([v1x, v1y], dtype=float)
v2 = np.array([0, v2y], dtype=float)

# -------------------------------
# Compute Collision Result
# -------------------------------
def compute_collision(m1, m2, v1, v2):
    Vcg = (m1 * v1 + m2 * v2) / (m1 + m2)
    u1 = v1 - Vcg
    u2 = v2 - Vcg

    angle = atan2(u1[1], u1[0])
    R = np.array([[np.cos(angle), np.sin(angle)],
                  [-np.sin(angle), np.cos(angle)]])
    
    uac1 = R @ u1
    uac2 = R @ u2

    wac2x = ((1 - m1 / m2) * uac2[0] + 2 * m1 / m2 * uac1[0]) / (1 + m1 / m2)
    wac1x = uac2[0] - uac1[0] + wac2x

    wac1 = np.array([wac1x, 0])
    wac2 = np.array([wac2x, 0])

    R_inv = np.linalg.inv(R)
    w1 = R_inv @ wac1 + Vcg
    w2 = R_inv @ wac2 + Vcg

    return w1, w2

w1, w2 = compute_collision(m1, m2, v1, v2)
alpha_1 = round(degrees(atan2(w1[1], w1[0])) / 10) * 10
alpha_2 = round(degrees(atan2(w2[1], w2[0])) / 10) * 10

x1_init = -v1 * (t_stop - tcoll)
x2_init = -v2 * (t_stop - tcoll)

x1_coll = x1_init + v1 * tcoll
x2_coll = x2_init + v2 * tcoll

# -------------------------------
# Set Up Figure
# -------------------------------
fig, ax = plt.subplots(figsize=(6, 6))
ax.set_xlim(-300, 300)
ax.set_ylim(-300, 300)
ax.set_xticklabels([])
ax.set_yticklabels([])
ax.set_aspect('equal')
ax.grid()

p1, = ax.plot([], [], 'ro')
p2, = ax.plot([], [], 'bo')
path1, = ax.plot([], [], 'r--', lw=1)
path2, = ax.plot([], [], 'b--', lw=1)
angle_text = ax.text(0.02, 0.02, '', transform=ax.transAxes)

traj1, traj2 = [], []

# -------------------------------
# Animation Functions
# -------------------------------
def init():
    p1.set_data([], [])
    p2.set_data([], [])
    path1.set_data([], [])
    path2.set_data([], [])
    angle_text.set_text('')
    return p1, p2, path1, path2, angle_text

def update(frame):
    t = frame * dt
    if t < tcoll:
        pos1 = x1_init + v1 * t
        pos2 = x2_init + v2 * t
    else:
        pos1 = x1_coll + w1 * (t - tcoll)
        pos2 = x2_coll + w2 * (t - tcoll)

    traj1.append(pos1.copy())
    traj2.append(pos2.copy())

    p1.set_data([scale * pos1[0]], [scale * pos1[1]])
    p2.set_data([scale * pos2[0]], [scale * pos2[1]])

    traj1_np = np.array(traj1)
    traj2_np = np.array(traj2)

    path1.set_data(scale * traj1_np[:, 0], scale * traj1_np[:, 1])
    path2.set_data(scale * traj2_np[:, 0], scale * traj2_np[:, 1])

    if abs(t - t_stop) < dt:
        angle_text.set_text(f"α₁ = {alpha_1}°, α₂ = {alpha_2}°")

    return p1, p2, path1, path2, angle_text

# -------------------------------
# Create and Display Animation
# -------------------------------
ani = FuncAnimation(fig, update, frames=int(t_stop / dt), init_func=init, blit=True, interval=50)
HTML(ani.to_jshtml())

7.6.1.2.22D elastic collision¶

Next, we consider an elastic collision between $m_1$ and $m_2$ , but now in a 2-dimensional setting.

In the widget below, the situation is animated. You can change the values of the initial velocity and masses. Can you (qualitatively) predict the outcome of the collision for a given set os parameters?

7.6.1.3Inelastic Collision¶

PArticle $m_1$ is moving over the $x$ -axis with unit velocity. Simultaneously, particle $m_2$ is moving over the $y$ -axis also with unit velocity. Both particles will collide in the origin. The collision is inelastic with restitution coefficient $e$ .

The widget below shows the trajectories of the particles and gives the velocities after the collision. Moreover, als the angle of the trajectories after the collision with the $x$ -axis is given.

Can you solve this problem for a few values of the restitution coefficient? The ‘easy ones’ are for $e=0$ .

Solution to Exercise 2

Given: the collision is completely inelastic. That means $e=0$ or in words: after the collision the two particles stick together and move as one particle. Thus, we have only one unknown velocity after the collision.

The problem is 1-dimensional and we can solve it by requiring conservation of momentum:

$$\begin{split}

\text{before ;;;;;;;} mv &+ M \cdot 0 = (m+M)U \text{ ;;;;;;; after} \ \Rightarrow U &= \frac{m}{m+M} v

\end{split}$$

7.6.2Exercises¶

7.6.2.1Answers¶

Solution to Exercise 4

$3m$ has velocity $-2v$ and $2m$ has velocity $3v$
Both particles have zero velocity.
$3m$ has velocity $-2/5 v$ and $2m$ has velocity $3/5 v$ .

Solution to Exercise 5

\vec{v}_{after} = \frac{2}{3}v \hat{x} + \frac{1}{3}v \hat{y}

(1)

\Delta E_{kin} = -\frac{2}{3}mv^2

(2)

Solution to Exercise 6

After each bounce, the tennis ball reaches half of the height it had before the bounce. Thus after two bounces, the ball reaches 25cm and with the third bounce only 12.5cm.

Solution to Exercise 7

We can consider the shooting as a collision. Bullets don’t bounce back, they penetrate a body. So the victim ‘gains’ maximum momentum if the bullet stays in the body. Then according to conservation of momentum, we have for this inelastic collision:

m_b v_b + M_v \cdot 0 = \left ( m_b + M_v \right ) U

(3)

Thus the velocity of the victim after being shot is:

U = \frac{m_b}{m_b + M_v} v_b

(4)

For the shooter a similar argument holds: before the shot, bullet & shooter have zero momentum. After firing, the bullet has velocity $v_b$ . Thus conservation of momentum requires:

0 = m_b v_b + M_s U_s

(5)

and we find for the velocity of the shooter:

U_s = -\frac{m_b}{M_s} v_b

(6)

Conclusion: as the mass of the bullet is negligible compared to that of a person both shooter and victim have similar velocities. As their mass is comparable, it is clear: from a physics point of view, if the victim is blown backward, than also the shooter is.

From the above we get, using $m_b \approx 10 \cdot 10^{-3}$ kg, $v_b \approx 500$ m/s and $M_v \approx 70$ kg:

U_v = \frac{m_b}{m_b + M_v} v_b \approx 7 cm/s

(7)

That is much too little to ‘knock’ someone over. Hollywood is good at ‘dramatic effects’, not so good at physics.

Classical Mechanics & Special Relativity for Starters

Collisions

Classical Mechanics & Special Relativity for Starters

Two Body Problem: Kepler revisited

7.6Exercises, examples & solutions

7.6.1Examples¶

7.6.1.1Newton’s Cradle¶

7.6.1.2Elastic Collision¶

7.6.1.2.11D elastic collision¶

7.6.1.2.22D elastic collision¶

7.6.1.3Inelastic Collision¶

7.6.2Exercises¶

7.6.2.1Answers¶