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

Worked Example 1: Ball and Bat Collision

Problem: A ball of mass 0.2 kg moves with velocity $-8$ m s$^{-1}$ horizontally towards a bat. The bat strikes it, reversing its direction, and the ball moves away at $12$ m s$^{-1}$.

Find: a) The impulse of the bat on the ball
b) The impulse of the ball on the bat

Solution:

Step 1: Apply the impulse equation for the ball.

We use: $I = mv - mu$

Taking the positive direction to the right:

• Initial velocity: $u = -8$ m s$^{-1}$ (moving left, hence negative)
• Final velocity: $v = 12$ m s$^{-1}$ (moving right, hence positive)
• Mass: $m = 0.2$ kg

$I = mv - mu$ $I = 0.2 \times 12 - 0.2 \times (-8)$ $I = 2.4 + 1.6 = 4.0 \text{ Ns}$

The impulse of the bat on the ball is 4.0 Ns to the right.

Step 2: Apply Newton's third law.

Newton's third law states that forces (and impulses) between two interacting objects are equal in magnitude but opposite in direction.

The impulse of the ball on the bat is 4.0 Ns in the opposite direction (to the left).

Exam tip: Always check your signs carefully. The change from $-8$ to $+12$ m s$^{-1}$ represents a large momentum change because the ball reverses direction.

Worked Example 2: Water Striking a Wall

Problem: Water flows horizontally from a pipe with cross-sectional area $0.02$ m$^2$ at a speed of $15$ m s$^{-1}$. It strikes a fixed vertical wall.

Find the force F on the wall. (Density of water = $1000$ kg m$^{-3}$)

Solution:

Step 1: Calculate the mass of water hitting the wall per second.

Volume of water per second = cross-sectional area $\times$ speed

$\text{Volume per second} = 0.02 \times 15 = 0.3 \text{ m}^3$

Mass per second = volume $\times$ density

$\text{Mass per second} = 0.3 \times 1000 = 300 \text{ kg}$

Step 2: Use the impulse equation to find the constant force.

The impulse equation is: $I = Ft = mv - mu$

Consider a time period of $t = 1$ second:

• Initial velocity of water: $u = 15$ m s$^{-1}$ (towards wall)
• Final velocity: $v = 0$ m s$^{-1}$ (water stops at wall)
• Mass in 1 second: $m = 300$ kg

$F \times 1 = 0 - 300 \times 15$ $F = -4500 \text{ N}$

The negative sign indicates the force on the water opposes its initial motion (the wall pushes back).

Step 3: Apply Newton's third law.

The force exerted by the wall on the water is $-4500$ N (opposing the water's motion).

By Newton's third law, the force exerted by the water on the wall is equal and opposite:

Force on the wall = 4500 N

Note: All the water's momentum is destroyed on impact. The water arrives with momentum and leaves with zero momentum in the horizontal direction.

Worked Example 3: Particle with Variable Force

Problem: A particle of mass 2 kg has an initial velocity $u$ of $3$ m s$^{-1}$. It experiences a force $F = 8t - 3t^2$ newtons for 3 seconds, where $0 \leq t \leq 3$.

Find its speed $v$ after 3 seconds.

Solution:

Step 1: Calculate the impulse using integration.

Since the force varies with time, we must integrate:

$I = \int_0^3 F \, dt = \int_0^3 (8t - 3t^2) \, dt$

Integrating term by term:

$I = \left[\frac{8t^2}{2} - \frac{3t^3}{3}\right]_0^3 = \left[4t^2 - t^3\right]_0^3$

Evaluating at the limits:

$I = (4 \times 9 - 27) - 0 = 36 - 27 = 9 \text{ Ns}$

Step 2: Use the impulse equation to find the final velocity.

Now we have the total impulse, we can find the velocity change:

$I = mv - mu$ $9 = 2v - 2 \times 3$ $9 = 2v - 6$ $2v = 15$ $v = 7.5 \text{ m s}^{-1}$

The final speed is 7.5 m s$^{-1}$

Exam tip: When forces vary with time, always use integration to find impulse. Don't try to use $I = F \times t$ with a variable force.

Worked Example 4: Ball Impacting a Plane

Problem: A 2 kg ball with velocity u = $3$i $- 4$j m s$^{-1}$ impacts a smooth plane that is parallel to the $x$-axis. The coefficient of restitution $e = 0.5$.

Calculate: a) The final velocity v = $v_x$i $+ v_y$j of the ball
b) The impulse I of the plane on the ball

Solution:

Step 1: Understand the situation.

The ball approaches with:

• Horizontal component: $3$ m s$^{-1}$ (parallel to the plane)
• Vertical component: $-4$ m s$^{-1}$ (towards the plane, downwards)

Since the plane is smooth, there is no friction. This means there's no impulse component parallel to the plane.

Step 2: Find the velocity component parallel to the plane.

With no impulse along the plane, momentum is conserved in the $x$-direction:

$\text{Initial momentum} = \text{Final momentum}$ $2 \times 3 = 2 \times v_x$ $v_x = 3 \text{ m s}^{-1}$

The horizontal velocity component remains unchanged.

Step 3: Find the velocity component perpendicular to the plane.

Apply Newton's law of restitution:

$e = \frac{\text{speed of separation}}{\text{speed of approach}}$

The speed of approach perpendicular to the plane is $4$ m s$^{-1}$ (taking magnitude).

$0.5 = \frac{v_y}{4}$ $v_y = 2 \text{ m s}^{-1}$

The ball bounces back with $v_y = 2$ m s$^{-1}$ (upwards, positive direction).

The final velocity is v = $3$i $+ 2$j m s$^{-1}$

Step 4: Calculate the impulse perpendicular to the plane.

The impulse I acts only perpendicular to the plane (in the $y$-direction).

Using the impulse equation in the $y$-direction:

$I = m v_y - m u_y$ $I = 2 \times 2 - 2 \times (-4)$ $I = 4 + 8 = 12 \text{ Ns}$

The impulse is 12 Ns in the positive $y$-direction.

As a vector: I = $12$j Ns

Exam tip: For smooth surfaces, remember that momentum is conserved parallel to the surface because there's no impulse component in that direction.