Section 5.2: Impulse-Momentum Theorem
The Impulse-Momentum Theorem states that the impulse applied to an object equals the change in its momentum:
\( \vec{J} = \Delta \vec{p} = m \vec{v}_f - m \vec{v}_i \)
This links force, time, and motion: applying a force over a time interval changes the momentum.
- Impulse \( \vec{J} = \vec{F} \Delta t \)
- Change in momentum \( \Delta \vec{p} = m \vec{v}_f - m \vec{v}_i \)
- Equating: \( \vec{F} \Delta t = m \vec{v}_f - m \vec{v}_i \)
Example 1
A 0.2 kg ball moving at 10 m/s is stopped by a player’s hand in 0.05 s. Find the average force applied.
Initial momentum: \( p_i = 0.2 \times 10 = 2 \text{ kg·m/s} \)
Final momentum: \( p_f = 0 \)
Impulse: \( J = p_f - p_i = -2 \text{ N·s} \)
Average force: \( F = J / \Delta t = -2 / 0.05 = -40 \text{ N} \)
Example 2
A 5 kg cart moving at 2 m/s is pushed with a constant 10 N force for 4 s. Find final velocity.
Impulse: \( J = F \Delta t = 10 \times 4 = 40 \text{ N·s} \)
Initial momentum: \( p_i = 5 \times 2 = 10 \text{ kg·m/s} \)
Final momentum: \( p_f = p_i + J = 10 + 40 = 50 \text{ kg·m/s} \)
Final velocity: \( v_f = p_f / m = 50 / 5 = 10 \text{ m/s} \)
Practice Problems
- A 3 kg object at 4 m/s is stopped in 0.2 s. Find the average force.
- A 0.5 kg ball changes velocity from 6 m/s to -4 m/s in 0.1 s. Find impulse and force.
- A 2 kg cart moving at 3 m/s is accelerated to 7 m/s by a force applied for 2 s. Find the force.
- A 1500 kg car slows from 20 m/s to 5 m/s in 10 s. Compute the braking force.
- A 0.25 kg baseball is hit from 30 m/s to 50 m/s in 0.02 s. Find the force applied by the bat.
- A 4 kg object at rest is pushed with 12 N for 3 s. Find final velocity.
- A hockey puck of mass 0.16 kg is accelerated from 2 m/s to 6 m/s in 0.5 s. Determine force.
- A 5 kg ball moving at 8 m/s is stopped in 0.1 s. Find impulse.
- A truck of mass 2000 kg increases velocity from 0 to 15 m/s in 5 s. Find average force.
- A 0.8 kg ball is struck to change velocity from -10 m/s to 6 m/s in 0.2 s. Compute impulse and force.