A 20g bullet moving at 200m/s hit a bag of sand and comes to rest in 0.011s.what is the momentum of the bullet just before hitting the bag? find the average force that stopped the bullet
To find the momentum of the bullet just before it hits the bag, we can use the formula:
Momentum (p) = mass (m) × velocity (v)
Given:
Mass (m) = 20g = 0.02 kg (converted from grams to kilograms)
Velocity (v) = 200 m/s
Using the formula, substitute the given values:
Momentum (p) = 0.02 kg × 200 m/s
Calculating this equation, we find:
Momentum (p) = 4 kg·m/s
Now, to find the average force that stopped the bullet, we can use the formula:
Force (F) = Change in momentum (Δp) / Time (t)
Given:
Change in momentum (Δp) = Final momentum - Initial momentum
Time (t) = 0.011s
Since the bullet comes to rest, the final momentum is 0 kg·m/s. Therefore, the change in momentum is equal to the initial momentum (4 kg·m/s).
Substituting the values into the formula:
Force (F) = 4 kg·m/s / 0.011s
Calculating this equation, we find:
Force (F) = 363.6 N
Therefore, the momentum of the bullet just before hitting the bag is 4 kg·m/s, and the average force that stopped the bullet is 363.6 N.