Question

example:
a vehicle travelling along a street skidded on an icy patch and hit a roadside barrier. the time interval for the impact was 0.955 s, during which the barrier exerted a force of 505 n. the magnitude of the change in momentum for the vehicle is __ __ kg·m/s.

example:
a vehicle experiences a change in momentum of +356 kg·m/s over a time interval of 5.81 s. the force on the vehicle during this time is +/- __, , , __ n.

example:
a 25 kg cart, initially at rest, is pushed ahead for 3.50 s until it has a velocity of 1.50 m/s. the magnitude of the force required to do this is __, , , __ n.

Explanation:

First Example (Change in Momentum)

Step1: Recall Impulse-Momentum Theorem

The impulse-momentum theorem states that the impulse (\(J\)) is equal to the change in momentum (\(\Delta p\)), and impulse is also given by \(J = F \cdot \Delta t\), where \(F\) is the force and \(\Delta t\) is the time interval. So, \(\Delta p = F \cdot \Delta t\).

Step2: Substitute Values

Given \(F = 505\ \text{N}\) and \(\Delta t = 0.955\ \text{s}\), we calculate \(\Delta p = 505\ \text{N} \times 0.955\ \text{s}\).

$$ \Delta p = 505 \times 0.955 = 482.275\ \text{kg·m/s} $$

Step1: Recall Impulse-Momentum Theorem

From \(J = \Delta p = F \cdot \Delta t\), we can solve for force \(F=\frac{\Delta p}{\Delta t}\).

Step2: Substitute Values

Given \(\Delta p = 356\ \text{kg·m/s}\) and \(\Delta t = 5.81\ \text{s}\), we calculate \(F=\frac{356}{5.81}\).

$$ F=\frac{356}{5.81}\approx 61.27\ \text{N} $$

Step1: Find Change in Momentum

The cart starts at rest (\(v_i = 0\)) and has a final velocity \(v_f = 1.50\ \text{m/s}\), mass \(m = 25\ \text{kg}\). The change in momentum \(\Delta p = m(v_f - v_i)=m v_f\) (since \(v_i = 0\)). So, \(\Delta p = 25\ \text{kg} \times 1.50\ \text{m/s}=37.5\ \text{kg·m/s}\).

Step2: Find Force Using Impulse-Momentum Theorem

Using \(F=\frac{\Delta p}{\Delta t}\), with \(\Delta t = 3.50\ \text{s}\), we get \(F=\frac{37.5}{3.50}\).

$$ F=\frac{37.5}{3.50}\approx 10.71\ \text{N} $$

Answer:

\(482\) (or more precisely \(482.28\) depending on rounding, but likely \(482\) or \(482.3\) if more decimal places are considered)

Second Example (Force from Change in Momentum)