A 110 kg rugby player running at 8.00 m/s collides head on with a padded goalpost, which exerts a backward force of 1.76 times ten to the fourth newtons on him for 5.50 times ten to the minus 2 seconds. What is his final velocity?

1. Taking the player's original direction as positive, the backward force gives a negative impulse. 2. That impulse is delta p equal to negative (1.76 times ten to the fourth N)(5.50 times ten to the minus 2 s), which is negative 968 kg m/s. 3. The player's final velocity is v_f equal to v_i plus delta p over m, which is 8.00 m/s plus (negative 968 kg m/s divided by 110 kg). 4. That gives 8.00 m/s minus 8.80 m/s, or negative 0.80 m/s, meaning he ends up moving backward at 0.80 m/s. 5. 8.80 m/s forward or backward would come from treating the change in velocity as the final velocity itself, without adding it to the original 8.00 m/s. _Source: OpenStax College Physics (CC BY 4.0), Ch 8 "Linear Momentum and Collisions", section 8.2 Impulse_