Let us suppose that we apply a force \( \vec{F} \) to an object for \( \Delta t \) seconds. What does the quantity \( \vec{F} \cdot \Delta t \) represent?
According to Newton's second law, the net force \( \vec{F} \) on any object is related to its acceleration \( \vec{a} \) and mass \( m \): \[ \vec{F} = m \vec{a} \] where acceleration is defined as \[ \vec{a} = \frac{\Delta \vec{v}}{\Delta t} = \frac{\vec{v_f} - \vec{v_i}}{\Delta t} \] \( \vec{v_f} \) is the final velocity and \( \vec{v_i} \) is the initial velocity. \( \Delta t \) is the time interval.
Substitute \( \vec{a} \) into Newton's second law: \[ \vec{F} = m \frac{\Delta \vec{v}}{\Delta t} \] Rearranging gives: \[ \vec{F} \, \Delta t = m \, \Delta \vec{v} = m (\vec{v_f} - \vec{v_i}) = m \vec{v_f} - m \vec{v_i} = \vec{p_f} - \vec{p_i} \] where \( \vec{p_f} \) is the final momentum and \( \vec{p_i} \) is the initial momentum.
The quantity \( \vec{F} \, \Delta t \) represents the change in momentum and is called the impulse of the force \( \vec{F} \) over time \( \Delta t \).
You need to stop a \( 0.4 \, \text{kg} \) ball moving at \( 20 \, \text{m/s} \). What force is required to stop the ball in:
Solution:
\[ F \, \Delta t = m \, \Delta v = m (v_f - v_i) \] We need to stop the ball, so \( v_f = 0 \), \( v_i = 20 \, \text{m/s} \) (assume positive direction). \[ F \, \Delta t = 0.4 \, (0 - 20) = -8 \, \mathrm{kg\cdot m/s} \] The negative sign indicates \( F \) opposes the motion. \[ F = -\frac{8}{\Delta t} \]
Stopping the same ball at \( 20 \, \text{m/s} \) requires a larger force as the time interval decreases.
A \( 2000 \, \text{kg} \) car traveling at \( 25 \, \text{m/s} \) is stopped in \( 10 \, \text{s} \) by a braking force \( F \). What is the magnitude of \( F \)?
Solution:
Initial velocity: \( v_i = 25 \, \text{m/s} \) (positive direction)
Final velocity: \( v_f = 0 \)
Initial momentum: \( p_i = m v_i = 2000 \times 25 = 50000 \, \mathrm{kg\cdot m/s} \)
Final momentum: \( p_f = 0 \)
\[
|F| \, \Delta t = |p_f - p_i| = |0 - 50000| = 50000
\]
\[
|F| = \frac{50000}{10} = 5000 \, \text{N}
\]
A football player kicks a \( 350 \, \text{g} \) ball (initially at rest), giving it a speed of \( 28 \, \text{m/s} \). Find the average force exerted if the impact time is \( 12 \, \text{ms} \).
Solution:
Mass: \( m = 0.35 \, \text{kg} \)
Initial velocity: \( v_i = 0 \)
Final velocity: \( v_f = 28 \, \text{m/s} \)
\[
|F| \, \Delta t = |p_f - p_i| = |0.35 \times 28 - 0| = 9.8
\]
\[
|F| = \frac{9.8}{12 \times 10^{-3}} \approx 817 \, \text{N}
\]
A \( 1 \, \text{kg} \) ball hits a wall at \( 10 \, \text{m/s} \) left and rebounds at \( 8 \, \text{m/s} \) right. What average force does the wall exert if contact lasts \( 0.1 \, \text{s} \)?
Solution:
Let right be positive.
\( v_i = -10 \, \text{m/s} \), \( v_f = 8 \, \text{m/s} \)
\[
F \, \Delta t = p_f - p_i = (1 \times 8) - (1 \times (-10)) = 18
\]
\[
F = \frac{18}{0.1} = 180 \, \text{N} \quad \text{(to the right)}
\]
Two trucks travel at the same speed. Truck A has twice the mass of truck B. Both apply equal braking forces. Which truck stops in a longer distance?
Solution:
Let \( m \) be truck B’s mass, so truck A’s mass is \( 2m \).
Initial speed: \( v_i \) (same for both).
Final speed: \( 0 \).
For each truck, \( F \, \Delta t = 0 - m v_i \) (with \( F \) negative).
Truck A: \( \Delta t_A = -\frac{2m v_i}{F} \)
Truck B: \( \Delta t_B = -\frac{m v_i}{F} \)
Since \( |\Delta t_A| > |\Delta t_B| \), truck A takes longer to stop and therefore requires a longer stopping distance (twice as long, assuming constant deceleration).