Question

1 impulse
$j = ft$
$p = mv$
problem 1. a 4 kg dog is running with a velocity of 6 m/s directly into the arms of jayson, completely stopping its momentum. if the collision happens in a time of $t = 0.5 s$, calculate the amount of force jayson used to stop the dog.

problem 2. a 3 kg turtle is sliding down a rock with a velocity of 2 m/s. the turtle collides into a tree that completely stops its momentum in 0.1 s. how much force was applied by the tree to stop the momentum of the turtle?

problem 3. a 100 kg buffalo is running towards a lake with a velocity of 7 m/s. when the buffalo runs into the lake it completely stops its momentum. if the lake applies a force of 550 n, determine the time taken during the collision.

problem 4. an acrobat with a mass of 25 kg falls with a velocity of 11 m/s towards a net below, completely stopping their momentum. if the force applied by the net is 150 n, determine how much time it takes for the collision to occur.

Explanation:

Problem 1

Step1: Calculate initial momentum

The formula for momentum is \( p = mv \). Given \( m = 4\space kg \) and \( v = 6\space m/s \), we substitute these values into the formula:
\( p = 4 \times 6 = 24\space kg\cdot m/s \)

Step2: Relate impulse and momentum

Impulse \( J \) is equal to the change in momentum. Since the dog comes to rest, the change in momentum \( \Delta p = p - 0 = 24\space kg\cdot m/s \). Impulse is also given by \( J = Ft \). So we can set \( Ft=\Delta p \).

Step3: Solve for force \( F \)

We know \( \Delta p = 24\space kg\cdot m/s \) and \( t = 0.5\space s \). Rearranging \( F=\frac{\Delta p}{t} \), we substitute the values:
\( F=\frac{24}{0.5}=48\space N \)

Step1: Calculate initial momentum

Using \( p = mv \), with \( m = 3\space kg \) and \( v = 2\space m/s \):
\( p = 3 \times 2 = 6\space kg\cdot m/s \)

Step2: Relate impulse and momentum

The change in momentum \( \Delta p = 6 - 0 = 6\space kg\cdot m/s \) (since the turtle stops). Impulse \( J = Ft \), so \( Ft=\Delta p \).

Step3: Solve for force \( F \)

Given \( t = 0.1\space s \), rearranging \( F=\frac{\Delta p}{t} \):
\( F=\frac{6}{0.1}=60\space N \)

Step1: Calculate initial momentum

Using \( p = mv \), with \( m = 100\space kg \) and \( v = 7\space m/s \):
\( p = 100 \times 7 = 700\space kg\cdot m/s \)

Step2: Relate impulse and momentum

The change in momentum \( \Delta p = 700 - 0 = 700\space kg\cdot m/s \). Impulse \( J = Ft \), so \( Ft=\Delta p \).

Step3: Solve for time \( t \)

Given \( F = 550\space N \), rearranging \( t=\frac{\Delta p}{F} \):
\( t=\frac{700}{550}=\frac{14}{11}\approx1.27\space s \) (rounded to two decimal places)

Answer:

The force Jayson used is \( \boldsymbol{48\space N} \)

Problem 2