Magnitude and Impulse on Person-please help explain, I am very confused.

Question

Erica (22 kg) and Danny (33 kg) are bouncing on a trampoline. Just as Erica reaches the highest point of her bounce, Danny is moving upward past her, and at that instant grabs a hold of her. If their speed just after he grabs her is 2.0 m/s, what was the magnitude of the impulse on Erica. (answer 44 N-s)

Answer 1

Hi Lauren

The impulse on a person or object is equal to the change in momentum. The momentum of an object is calculated by multiplying its mass by its velocity.

So, first, you need to find the momentum of Erica just before Danny grabs her:

\(P_{\text{Erica\_initial}} = m_{\text{Erica}} \cdot v_{\text{Erica\_initial}}\)

where:

\(m_{\text{Erica}} = 22 \, \text{kg}\) (mass of Erica) and

\(v_{\text{Erica\_initial}}\) is her velocity just before Danny grabs her.

Since the trampoline is at the highest point of the bounce, Erica's velocity is momentarily \(0 \, \text{m/s}\) at the top. Now, calculate her initial momentum:

\(P_{\text{Erica\_initial}} = 22 \, \text{kg} \cdot 0 \, \text{m/s} = 0 \, \text{Ns}\).

Now, you need to find the final momentum of Erica just after Danny grabs her. This is given as \(2.0 \, \text{m/s}\).

\(P_{\text{Erica\_final}} = m_{\text{Erica}} \cdot v_{\text{Erica\_final}}\)

\(P_{\text{Erica\_final}} = 22 \, \text{kg} \cdot 2.0 \, \text{m/s} = 44 \, \text{Ns}\).

Now, you can calculate the magnitude of the impulse on Erica, which is the change in momentum:

\(\text{Impulse} = |P_{\text{Erica\_final}} - P_{\text{Erica\_initial}}| = |44 \, \text{Ns} - 0 \, \text{Ns}| = 44 \, \text{Ns}\).

So, the magnitude of the impulse on Erica is \(44 \, \text{Ns}\).

I hope this helps out