Question

impulse = change in momentum: ( ft = delta mv )
part a
how much impulse stops a 45-kg carton sliding at 5.2 m/s when it meets a rough surface?
express your answer with the appropriate units.
( i = ) value units
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Explanation:

Step1: Recall the impulse - momentum theorem

The impulse \( I \) is equal to the change in momentum \( \Delta p \), and the formula is \( I=\Delta mv \). The initial velocity \( v_i = 5.2\ m/s \) and the final velocity \( v_f=0\ m/s \) (since the carton stops). The mass \( m = 45\ kg \). The change in velocity \( \Delta v=v_f - v_i=0 - 5.2=- 5.2\ m/s \), but we are interested in the magnitude of the impulse (or we can consider the absolute value for the stopping impulse).

Step2: Calculate the impulse

Using the formula \( I = m\Delta v \), substitute \( m = 45\ kg \) and \( \Delta v=5.2\ m/s \) (taking the magnitude of the change in velocity as the carton comes to rest). So \( I=m\times|v_f - v_i|=45\ kg\times5.2\ m/s \).
\[
I = 45\times5.2=234\ kg\cdot m/s
\]

Answer:

The impulse is \( 234\ kg\cdot m/s \) (or \( 234\ N\cdot s \) since \( 1\ N\cdot s = 1\ kg\cdot m/s \)).