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Video: Finding the Resistance of Barrier to a Penetrating Bullet

Ed Burdette

A bullet of mass 63 g was fired towards a fixed barrier at 80 m/s. Given that it penetrated 5 cm into the barrier before it stopped, find the resistance of the barrier to the bullet’s motion.


Video Transcript

A bullet of mass 63 grams was fired towards a fixed barrier at 80 meters per second. Given that it penetrated five centimeters into the barrier before it stopped, find the resistance of the barrier to the bullet’s motion.

We’re given the mass of this bullet as 63 grams, which we’ll call 𝑚. We’re also told that its initial speed, that is its speed right before it hits the barrier, is 80 meters per second. We’ll call that speed 𝑣 sub 𝑖. The distance the bullet penetrated into the barrier before stopping, five centimeters, we’ll call 𝑑. We want to solve for the resistance the barrier provides against the bullet’s motion. That resistance will be expressed as a force in newtons. And we’ll call it 𝐹. To start on our solution, let’s recall Newton’s second law of motion. The second law tells us that the net force acting on an object is equal to the mass of that object times its acceleration. If we also recall that acceleration is equal to the change in velocity over the change in time, then for our scenario, we can write that the resistive force on the bullet is equal to the bullet’s mass multiplied by its initial speed, right before it reaches the barrier, minus its final speed, all divided by the time it takes for the bullet to be slowed down.

In the problem statement, we’re told the mass of the bullet as well as its initial speed right before it reaches the barrier. Its final speed, 𝑣 sub 𝑓, will be zero because we know the bullet comes to rest. This means, if we can solve for the time 𝑡 it takes for the barrier to slow the bullet down to a stop, we can solve for the resistive force 𝐹. If we draw a diagram of our bullet being slowed down and stopped by our barrier, we can see that as soon as the bullet makes contact with the barrier, the barrier exerts a resistive force on the bullet’s motion. The bullet accelerates; that is, loses speed. And we’re going to assume that acceleration is constant.

Because we’ve made that assumption, that means that we can now apply a set of relations, called the kinematic equations, to the motion of our bullet as it slows to a stop. These four separate equations, each describe the motion of an object under constant acceleration. Since in our problem statement we’re told the bullet’s initial speed, we can infer its final speed. And we’re also told the distance that it travels while it comes to a stop. And we want to solve for the time 𝑡 that this all takes. We see that the fourth kinematic equation can be useful to us. Since for our scenario, 𝑣 sub 𝑓 — the final speed of the bullet — is zero, the equation reduces to 𝑑 equals 𝑣 sub 𝑖 over two times 𝑡, or 𝑡 equals two times 𝑑 over 𝑣 sub 𝑖.

We can take this expression for the time 𝑡 and substitute it into our equation for 𝐹. When we do, the equation becomes 𝑚 times 𝑣 sub 𝑖 squared over two times 𝑑. We’re given the mass 𝑚, the bullet’s initial speed, and the distance it penetrates into the barrier 𝑑. So we’re ready to plug in and solve for 𝐹. When we do, we’re careful to convert our mass into units of kilograms and our distance 𝑑 into units of meters. And all this is to ensure consistent units across this calculation.

When we enter these values on our calculator, we find that 𝐹 is 4032 newtons. That’s the resistive force of the barrier against the bullet’s motion.