What is the average force on Superman's chest when hit by 3g bullets at 500 m/s?

In summary, we are asked to calculate the magnitude of the average force on Superman's chest when he is sprayed with 3 g bullets at a rate of 100 bullets per minute, each traveling at a speed of 500 m/s and rebounding straight back. Using the equation F = Δp/Δt, we can find the average force by considering the change in momentum, which is equal to the product of mass and velocity. Taking into account the rebound, we can calculate the change in velocity to be 1000 m/s. By unit analysis, we can determine that the units for the average force must be in newtons, leading us to use the equation F = m*a. Ultimately, we can solve for the average
  • #1
Shatzkinator
53
0

Homework Statement


Suppose a gangster sprays Superman's chest with 3 g bullets at the rate of 100 bullets/min, and the speed of each bullet is 500 m/s. Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest?

Homework Equations


J = FavgT
p = mv
J = pf - pi

The Attempt at a Solution


I don't really understand how to get the delta T. I got J = -3 Ns from equation 3, don't know where to go from there.
 
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  • #2
You have 300 grams of bullets a minute hitting at 5 grams a sec on average.

Taking rebound into account you have a change in velocity of 1000 m/s then don't you have 5k grams/s*m/s? That looks like Newtons to me.
 
  • #3
LowlyPion said:
You have 300 grams of bullets a minute hitting at 5 grams a sec on average.

Taking rebound into account you have a change in velocity of 1000 m/s then don't you have 5k grams/s*m/s? That looks like Newtons to me.

hmm interesting. So how would I know to play with the units rather than using typical linear momentum equations?
 
  • #4
Shatzkinator said:
hmm interesting. So how would I know to play with the units rather than using typical linear momentum equations?

Insight. Regardless of the equations you must end with the same units they are asking. Units are often a guide to the answer. But it's not like I'm suggesting you not use the equations, so much as suggesting you understand what the equations are relating to begin with.

F = m*a = m*Δv/Δt = Δmv/Δt = Δp/Δt
 

1. What is a collision in physics?

A collision in physics refers to the event when two or more objects come into contact with each other and exert forces on each other. This can result in a change in the objects' motion and/or deformation of the objects involved.

2. What is impulse in physics?

In physics, impulse is defined as the product of force and time. It is a measure of the change in momentum of an object during a collision or interaction with another object. Mathematically, impulse is represented by the equation I = FΔt, where F is the force applied and Δt is the duration of the force.

3. How is impulse related to momentum?

Impulse and momentum are directly related to each other. According to the impulse-momentum theorem, the impulse applied to an object is equal to the change in its momentum. This means that if the impulse is increased, the change in momentum will also increase, and vice versa.

4. What are some real-life examples of collisions and impulse?

Collisions and impulse can be observed in various real-life situations. Some examples include a car crash, a ball bouncing off a wall, a bat hitting a ball, a person catching a ball, and a person jumping on a trampoline. In all of these examples, there is a transfer of momentum and/or a change in motion due to the collision and impulse involved.

5. How can we calculate the impulse during a collision?

The impulse during a collision can be calculated using the equation I = FΔt, where F is the average force applied during the collision and Δt is the duration of the collision. This can be measured using force sensors and timing devices. Additionally, the impulse can also be calculated by integrating the force-time graph during the collision.

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