Question

question 3 of 25
a boat of mass 450 kg drifts along a river at a speed of 10 m/s to the north.
what impulse is required to decrease the speed of the boat to 6 m/s to the
north?

a. 1800 kg·m/s south
b. 1800 kg·m/s north
c. 2700 kg·m/s south
d. 2700 kg·m/s north

Explanation:

Step1: Recall the impulse - momentum theorem

The impulse - momentum theorem states that \(J=\Delta p = m\Delta v\), where \(J\) is the impulse, \(\Delta p\) is the change in momentum, \(m\) is the mass of the object, and \(\Delta v\) is the change in velocity.

Step2: Calculate the change in velocity

The initial velocity \(v_{i}=10\ m/s\) (north) and the final velocity \(v_{f} = 6\ m/s\) (north). The change in velocity \(\Delta v=v_{f}-v_{i}=6 - 10=- 4\ m/s\) (the negative sign indicates the direction is south).

Step3: Calculate the impulse

Given \(m = 450\ kg\) and \(\Delta v=-4\ m/s\). Using \(J = m\Delta v\), we have \(J=450\times(-4)=-1800\ kg\cdot m/s\). The negative sign indicates the direction of the impulse is south.

Answer:

A. \(1800\ kg\cdot m/s\) south