Collisions (AQA A-Level Further Maths): Revision Notes

Worked Example: Direct Collision Between Two Particles

Problem: A particle P with mass 4 kg and velocity 20 m s$^{-1}$ collides directly with particle Q with mass 12 kg travelling at 4 m s$^{-1}$ in the opposite direction. Given $e = \frac{1}{2}$, calculate the velocities $v_1$ and $v_2$ after impact.

Solution:

Step 1: Calculate the speed of approach.

Taking the positive direction as the direction P is moving initially:

• P has velocity $u_1 = 20$ m s$^{-1}$
• Q has velocity $u_2 = -4$ m s$^{-1}$ (opposite direction)

Speed of approach $= u_1 - u_2 = 20 - (-4) = 24$ m s$^{-1}$

Step 2: Write the momentum equation.

$4 \times 20 + 12 \times (-4) = 4v_1 + 12v_2$

$80 - 48 = 4v_1 + 12v_2$

$v_1 + 3v_2 = 8 \quad \text{...(1)}$

Step 3: Write Newton's equation.

$\frac{v_2 - v_1}{24} = \frac{1}{2}$

$v_2 - v_1 = 12 \quad \text{...(2)}$

Step 4: Solve equations (1) and (2) simultaneously.

Adding equations (1) and (2): $4v_2 = 20$ $v_2 = 5 \text{ m s}^{-1}$

Substituting back into equation (1): $v_1 = 8 - 3 \times 5 = 8 - 15 = -7 \text{ m s}^{-1}$

Answer: P moves at 7 m s$^{-1}$ in the opposite direction to its initial motion, and Q moves at 5 m s$^{-1}$ in the original positive direction. Both particles change direction after impact.

Worked Example: Oblique Impact with a Plane

Problem: A ball strikes a smooth fixed plane with velocity $\mathbf{u} = 20\sqrt{3}\mathbf{i} - 20\mathbf{j}$ m s$^{-1}$. The coefficient of restitution is $e = 0.8$. Find the final velocity $\mathbf{v} = v_x\mathbf{i} + v_y\mathbf{j}$ after impact and the angle $\theta$ it makes with the plane.

Solution:

Step 1: Identify components parallel and perpendicular to the plane.

The plane is horizontal, so:

• Parallel component: $u_x = 20\sqrt{3}$ m s$^{-1}$
• Perpendicular component: $u_y = 20$ m s$^{-1}$ (magnitude, approaching plane)

Step 2: Apply momentum conservation parallel to plane.

Since the plane is smooth, the horizontal component is unchanged:

$v_x = u_x = 20\sqrt{3} \text{ m s}^{-1}$

Step 3: Apply Newton's experimental law perpendicular to plane.

$\frac{v_y}{u_y} = e$

$\frac{v_y}{20} = 0.8$

$v_y = 20 \times 0.8 = 16 \text{ m s}^{-1}$

Step 4: Write the final velocity.

$\mathbf{v} = 20\sqrt{3}\mathbf{i} + 16\mathbf{j} \text{ m s}^{-1}$

Step 5: Calculate the angle using trigonometry.

$\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right) = \tan^{-1}\left(\frac{16}{20\sqrt{3}}\right) = 24.8°$

Answer: The ball rebounds with velocity $20\sqrt{3}\mathbf{i} + 16\mathbf{j}$ m s$^{-1}$ at an angle of approximately 24.8° to the plane.

Worked Example: Multiple Collisions with a Wall

Problem: Two spheres P (mass 1 kg) and Q (mass 3 kg) lie on a smooth horizontal plane with line PQ at right angles to a vertical wall. P moves at 10 m s$^{-1}$ and collides directly with Q, which is initially at rest. Q then hits the wall and rebounds. The coefficient of restitution between the spheres is 0.4 and between Q and the wall is $e$. Show that P and Q collide again if $e > \frac{1}{7}$.

Solution:

First impact (P colliding with Q):

Step 1: Draw a diagram showing velocities before and after first impact.

Before: P at 10 m s$^{-1}$, Q at rest. After: P at $v_1$, Q at $v_2$.

Step 2: Check for external forces.

There are no external forces during this collision, so momentum is conserved.

Step 3: Write equations.

Momentum equation: $1 \times 10 + 3 \times 0 = 1 \times v_1 + 3 \times v_2$ $10 = v_1 + 3v_2 \quad \text{...(1)}$

Newton's equation: $\frac{v_2 - v_1}{10} = 0.4$ $v_2 - v_1 = 4 \quad \text{...(2)}$

Step 4: Solve simultaneously.

From equations (1) and (2): $v_1 = -0.5 \text{ m s}^{-1}$ $v_2 = 3.5 \text{ m s}^{-1}$

P is now moving away from the wall, and Q is moving towards the wall.

Second impact (Q colliding with wall):

Step 1: Draw another diagram.

Before: Q at 3.5 m s$^{-1}$ towards wall. After: Q at $v_3$ away from wall.

Step 2: Note that momentum is not conserved because the wall exerts an external force on Q.

Step 3: Write Newton's equation only.

$\frac{v_3}{3.5} = e$ $v_3 = 3.5e$

P and Q are now both moving away from the wall.

Condition for re-collision:

For P and Q to collide again, Q must be moving faster than P: $v_3 > v_1$ $3.5e > 0.5$ $e > \frac{0.5}{3.5} = \frac{1}{7}$

Therefore, P and Q collide again if $e > \frac{1}{7}$.

Worked Example: Three Particles with Two Successive Collisions

Problem: Particles P, Q, and R have masses 3 kg, 2 kg, and 1 kg with velocities 6 m s$^{-1}$, 4 m s$^{-1}$, and 2 m s$^{-1}$ respectively. They move in a straight line in this order. P strikes Q with perfect elasticity ($e = 1$). A second collision between Q and R has coefficient of restitution $e$. Find the possible values of $e$ if there are no more collisions.

Solution:

First collision (P on Q):

Initially: P at 6 m s$^{-1}$, Q at 4 m s$^{-1}$, R at 2 m s$^{-1}$. After first impact: P at $u$, Q at $v$, R still at 2 m s$^{-1}$.

Momentum equation: $3 \times 6 + 2 \times 4 = 3u + 2v$ $3u + 2v = 26 \quad \text{...[1]}$

Newton's equation (with $e = 1$): $\frac{v - u}{6 - 4} = 1$ $v - u = 2 \quad \text{...[2]}$

Solving [1] and [2]: $u = 4.4 \text{ m s}^{-1}$ $v = 6.4 \text{ m s}^{-1}$

Second collision (Q on R):

After second impact: Q at $w$, R at $x$.

Momentum equation: $2 \times 6.4 + 1 \times 2 = 2w + x$ $2w + x = 14.8 \quad \text{...[3]}$

Newton's equation: $\frac{x - w}{6.4 - 2} = e$ $x - w = 4.4e \quad \text{...[4]}$

From [3] and [4]: $3w = 14.8 - 4.4e$

Conditions for no more collisions:

For no further collisions, we need $w > u$ (so Q doesn't catch P) and $x > w$ (so R stays ahead of Q).

Condition 1: $w > u$ $\frac{14.8 - 4.4e}{3} > 4.4$ $14.8 - 4.4e > 13.2$ $4.4e < 1.6$ $e < \frac{1.6}{4.4} = \frac{4}{11}$

Condition 2: $x > w$ From [4]: $x - w = 4.4e > 0$, so $e > 0$

Answer: Combining both conditions: $0 < e < \frac{4}{11}$