Momentum and Impulse (VCE SSCE Physics): Revision Notes

Worked Example: Calculating Momentum

Question: Calculate the momentum of a 2450 kg car driving at 81 km h$^{-1}$ north.

Solution:

First, convert the velocity to SI units: $\frac{81}{3.6} = 22.5 \text{ m s}^{-1}$

Now apply the formula: $p = mv$ $p = (2450)(22.5)$ $p = 5.51 \times 10^4 \text{ kg m s}^{-1} \text{ north}$

Worked Example: Collision Between Two People

Question: In a physical education class, Mark (86 kg) and Max (62 kg) hold inflatable fit balls and run towards each other. Mark runs at 4.35 m s$^{-1}$ north, and Max runs at 5.55 m s$^{-1}$ south. After the collision, Mark moves at 3.64 m s$^{-1}$ south. Calculate Max's velocity after the collision.

Solution:

Let north be positive. This means south is negative.

Before the collision:

• Mark: $u_1 = +4.35$ m s$^{-1}$
• Max: $u_2 = -5.55$ m s$^{-1}$

After the collision:

• Mark: $v_1 = -3.64$ m s$^{-1}$
• Max: $v_2 = ?$

Apply conservation of momentum: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ $(86)(4.35) + (62)(-5.55) = (86)(-3.64) + (62)v_2$ $v_2 = 5.53 \text{ m s}^{-1}$

The positive value indicates Max is moving north.

Worked Example: Sticky Collision

Question: A 200 g toy train moves at 1.22 m s$^{-1}$ south when a 150 g piece of plasticine is thrown at it, travelling at 2.45 m s$^{-1}$ south. The plasticine sticks to the train. Calculate the velocity of the combined system after the collision.

Solution:

Convert masses to SI units: $m_1 = 0.20$ kg, $m_2 = 0.150$ kg

Let south be positive.

$m_1u_1 + m_2u_2 = m_3v_3$ $(0.2)(1.22) + (0.15)(2.45) = (0.2 + 0.15)v_3$ $v_3 = 1.75 \text{ m s}^{-1}$

The positive value indicates the direction is south.

Worked Example: Explosive Collision

Question: Shen rides a bicycle at 18 km h$^{-1}$ west when he throws a 500 g football at 36 km h$^{-1}$ west. Calculate Shen's final velocity, given that he and the bicycle have a combined mass of 59 kg.

Solution:

Convert to SI units:

• $u_1 = \frac{18}{3.6} = 5$ m s$^{-1}$
• $u_2 = \frac{36}{3.6} = 10$ m s$^{-1}$
• $m_2 = 0.500$ kg

Let west be positive.

$m_1u_1 = m_2v_2 + m_3v_3$ $(59 + 0.500)(5) = (10)(0.500) + (59)v_3$ $v_3 = 4.96 \text{ m s}^{-1} \text{ west}$

Worked Example: Calculating Change in Momentum

Question: An archer fires a 32.0 g arrow at 79.0 m s$^{-1}$ east towards a stationary target. When the arrow hits the target, it takes 0.1 ms to stop. Calculate the change in momentum of the arrow.

Solution:

Convert to SI units: $m = 0.032$ kg

Let east be positive.

$\Delta p = mv - mu$ $\Delta p = 0.032 \times 0 - 0.032 \times 79.0$ $\Delta p = -2.53 \text{ kg m s}^{-1}$

The negative sign indicates the direction of the change in momentum is west.

Worked Example: Impulse

Question: A driver travels at 54 km h$^{-1}$ north, then slows to 9 km h$^{-1}$ over a speed bump. The braking takes 3.45 s, and the car and its contents have a mass of 1650 kg.

a) Calculate the change in momentum of the car.

b) Calculate the average force on the car as it slowed down, including direction.

Solution:

a) Convert to SI units: $u = \frac{54}{3.6} = 15 \text{ m s}^{-1}, \quad v = \frac{9}{3.6} = 2.5 \text{ m s}^{-1}$

Let north be positive.

$\Delta p = mv - mu$ $\Delta p = (1650)(2.5) - (1650)(15)$ $\Delta p = -2.06 \times 10^4 \text{ kg m s}^{-1}$

The negative sign indicates the direction is south.

b) Using impulse formula: $\Delta p = I = F_{\text{av}}\Delta t$ $F_{\text{av}} = \frac{-2.06 \times 10^4}{3.45}$ $F_{\text{av}} = -5.98 \times 10^3 \text{ N}$

The negative sign indicates the direction is south.

Worked Example: Inelastic Collision

Question: Two carts on a track, each with mass 0.150 kg, collide head-on at 4.5 m s$^{-1}$. After the collision, they move away from each other at 1.25 m s$^{-1}$.

a) Show that momentum is conserved.

b) Show that this collision is inelastic.

Solution:

a) Let movement to the right be positive.

$\sum p_{\text{before}} = (4.5)(0.15) + (-4.5)(0.15) = 0$ $\sum p_{\text{after}} = (1.25)(0.15) + (-1.25)(0.15) = 0$

Since $\sum p_{\text{before}} = \sum p_{\text{after}}$, momentum is conserved.

b) Note that kinetic energy is a scalar quantity, so there are no negative signs.

$\sum E_{k,\text{before}} = \frac{1}{2}(0.15)(4.5)^2 + \frac{1}{2}(0.15)(4.5)^2 = 3.04 \text{ J}$

$\sum E_{k,\text{after}} = \frac{1}{2}(0.15)(1.25)^2 + \frac{1}{2}(0.15)(1.25)^2 = 0.234 \text{ J}$

Since $\sum E_{k,\text{before}} > \sum E_{k,\text{after}}$, this collision is inelastic.

Worked Example: Constant Force Graph

Question: A box is pushed with a force of 75 N for 15 s. Calculate the impulse given to the box.

Solution:

The impulse equals the area under the graph: $I = 75 \times 15$ $I = 1.13 \times 10^3 \text{ N s}$

Worked Example: Changing Force

Question: A car is accelerated by an initial force of 35 kN, which decreases at a constant rate to zero over 8 s. Calculate the impulse given to the car.

Solution:

Convert to SI units: $F = 35 \times 10^3$ N

The area under the graph forms a triangle: $I = \frac{1}{2}bh$ $I = \frac{1}{2}(8)(35 \times 10^3)$ $I = 1.40 \times 10^5 \text{ N s}$

Worked Example: Complex Graph

Question: A person pushes a 30 kg sled through snow. The force-time graph is shown below. Calculate the change in momentum given to the sled in the 20 s journey. Assume friction is negligible.

Solution:

Recall that impulse equals change in momentum.

Break the graph into a rectangle and triangle: $I = \frac{1}{2}(4)(27.5) + (16)(27.5)$ $I = 495 \text{ N s}$

Alternatively, use the trapezium formula: $I = \frac{1}{2}(20 + 16)(27.5) = 495 \text{ N s}$

Worked Example: Combining Impulse and Momentum

Question: Scientists test a new material for a super high bounce ball. The ball has mass 10 g and is fired horizontally north at 17.5 m s$^{-1}$ into a wall. A detector measures the force on the ball over time.

a) Calculate the average force on the ball.

b) Calculate the final velocity of the ball after the collision.

Solution:

a) First, calculate the impulse from the area under the force-time graph: $I = \frac{1}{2}(10 \times 10^{-2})(5)$ $I = 0.25 \text{ N s}$

Then find average force: $F_{\text{av}} = \frac{I}{\Delta t}$ $F_{\text{av}} = \frac{0.25}{10 \times 10^{-2}}$ $F_{\text{av}} = 2.50 \text{ N}$

b) Let south be positive. Use impulse equals change in momentum: $I = \Delta p = mv - mu$ $v = \frac{I + mu}{m}$ $v = \frac{0.25 + (0.01)(-17.5)}{0.01}$ $v = 7.5 \text{ m s}^{-1}$

The positive answer indicates the direction is south.