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2. a mastery check: impulse-momentum equation version a pass revise 1. …

Question

  1. a mastery check: impulse-momentum equation

version a
pass
revise

  1. complete the table.
  1. kit the cat (5 kg) is running at 12 m/s when she starts sliding across a smooth floor. friction provides a 20 n stopping force.
  • what is kits change in momentum?

show your work.

  • how long does it take kit to stop moving?

show your work.

  1. matti (40 kg) is running after kit at a speed of 6 m/s, when he starts sliding across the same floor. friction applies a 100n stopping force.
  • what impulse has been applied after 2 seconds?
  • how much has mattis speed changed after 2 seconds?
  • how fast is matti going after 2 seconds?

Explanation:

2. What is Kit's change in momentum?

Step1: Determine the initial and final momentum

The formula for momentum is \(p = mv\). Kit's mass \(m = 6\space kg\), initial velocity \(v_{i}=12\space m/s\), and final velocity \(v_{f} = 0\space m/s\) (since she stops).
The initial momentum \(p_{i}=mv_{i}=6\times12 = 72\space kg\cdot m/s\), and the final momentum \(p_{f}=mv_{f}=6\times0=0\space kg\cdot m/s\).
The change in momentum \(\Delta p=p_{f}-p_{i}=0 - 72=- 72\space kg\cdot m/s\).

Step2: Use the impulse - momentum theorem

The impulse - momentum theorem states that \(F\Delta t=\Delta p\). But if we just calculate using \(p = mv\) (as above), we can also note that \(\Delta p=m\Delta v=m(v_{f}-v_{i})\)

2. How long does it take Kit to stop moving?

Step1: Use the impulse - momentum theorem \(F\Delta t=\Delta p\)

We know \(F=- 20\space N\) (negative because it's a stopping force), \(\Delta p=-72\space kg\cdot m/s\)
From \(F\Delta t=\Delta p\), we can solve for \(\Delta t\). Rearranging the formula gives \(\Delta t=\frac{\Delta p}{F}\)
Substitute the values: \(\Delta t=\frac{-72}{-20}=3.6\space s\)

3. What impulse has been applied after 2 seconds?

Step1: Use the formula for impulse \(J = F\Delta t\)

Given \(F=-100\space N\) (stopping force) and \(\Delta t = 2\space s\)
\(J=F\Delta t=-100\times2=-200\space N\cdot s\)

3. How much has Matti's speed changed after 2 seconds?

Step1: Use the impulse - momentum theorem \(J=\Delta p=m\Delta v\)

We know \(J=-200\space N\cdot s\), \(m = 40\space kg\)
From \(J=m\Delta v\), we can solve for \(\Delta v\). Rearranging gives \(\Delta v=\frac{J}{m}\)
Substitute the values: \(\Delta v=\frac{-200}{40}=-5\space m/s\)

3. How fast is Matti going after 2 seconds?

Step1: Use the formula \(v = v_{i}+\Delta v\)

Initial velocity \(v_{i}=6\space m/s\), \(\Delta v=-5\space m/s\)
\(v=6+( - 5)=1\space m/s\)

Answer:

  • Kit's change in momentum: \(-72\space kg\cdot m/s\)
  • Time for Kit to stop: \(3.6\space s\)
  • Impulse on Matti after 2 seconds: \(-200\space N\cdot s\)
  • Matti's speed change: \(-5\space m/s\)
  • Matti's speed after 2 seconds: \(1\space m/s\)