a 72 kg man on roller skates, initially at rest, fires a pistol due west. if the 35 g bullet travels at 270 m/s, find the velocity (including direction) of the man.
To find the velocity of the man after firing the pistol, we can use the law of conservation of momentum. According to the law of conservation of momentum, the total momentum before the event is equal to the total momentum after the event.
The momentum of an object is calculated by multiplying its mass by its velocity. Therefore, the momentum of the man before firing the pistol is 0 kg·m/s (since he is initially at rest).
Let's denote the initial velocity of the bullet as "v_bullet" and the final velocity of the man as "v_man."
Since momentum is a vector quantity, and the man and the bullet are traveling in opposite directions, we can write the equation:
(mass of the bullet) * (velocity of the bullet) = (mass of the man) * (velocity of the man)
Plugging in the values:
(0.035 kg) * (270 m/s) = (72 kg) * (v_man)
Now, solve for v_man:
v_man = (0.035 kg * 270 m/s) / 72 kg = 0.13125 m/s
Therefore, the velocity of the man, including direction, is 0.13125 m/s in the opposite direction of the bullet, which is due east.