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

Answer:

$2.4 N\cdot s$.

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:

$\Sigma F \cdot \Delta t = (0.40 kg) \cdot [- 2 \frac{m}{s} - 4 \frac{m}{s} ]\\\Sigma F \cdot \Delta t = - 2.4 N \cdot s$

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