Question

A 0.40-kg object is traveling to the right (in the positive direction) with a speed of 4.0 m/s. After a 0.20 s collision, the object is traveling to the left at 2.0 m/s. What is the magnitude of the impulse (in N-s) acting on the object

Answer 1

`2.4 N\cdot s`.

Step-by-step explanation:

The equation that models the travel of the object is:

`m \cdot v{o} + \Sigma F \cdot \Delta t = m \cdot v_(f)`

The impulse is:

`\Sigma F \cdot \Delta t = m \cdot (v_(f) - v_(o))`

By replacing terms:

`\Sigma F \cdot \Delta t = (0.40 kg) \cdot [- 2 (m)/(s) - 4 (m)/(s) ]\\\Sigma F \cdot \Delta t = - 2.4 N \cdot s`

The magnitude of the impulse acting on the object is
`2.4 N\cdot s`.