Understanding $F=ma$

I know the formula F=ma, but I truly need to understand its implications.

Take this scenario for example, we have two bullets exactly the same weight and size (identical to each other). Lets assume their weights to be 100 grams each.

We also have two packets filled with, say rubber, again exactly the same weight and size. Lets assume their weights to be 2kgs each.

Now both these bullets are fired into the packets. Both the bullets hit the packets at the exact same velocity. However, the bullets acceleration were different. The first bullet had an acceleration of say 5m/s2, and the second bullet had an acceleration of 10m/s2. The bullets travel different distances and for a different period of time, so that when they impact the packets, they have the exact same velocity, say 100m/s.

In such a scenario, how will the impact on the two packets be different?

Also, if the force applied is only dependent on mass and acceleration, will the damage to the packet be same even if the bullets hit the packet at a much lesser velocity, say 10m/s only.

And would this scenario be any different in vacuum and in an atmosphere?

You're confusing cause and effect.

The better equation here is $A={F \over M}$

When a force is applied to an object, the object is accelerated. The object gains kinetic energy from the force, but it doesn't have a property of acceleration.

The gun accelerates the bullets with some acceleration for some time period. One gun could accelerate the bullet slowly down a long barrel, while the second gun accelerates the bullet quickly down a short barrel, and end up with the same velocity. But once both bullets leave the guns with the same velocity, the acceleration is irrelevant to how they impact.

The same equation works on impact. It's a bit more complex, because we have to deal with elasticity and shear strength, but basically the rubber packets push back with some force, causing them to decelerate. With a given starting velocity, a higher-mass bullet will tend to penetrate further since the same force will decelerate it less.

Similarly, a lower velocity will result in less force being applied to both the bullets and the packets. But the acceleration up front is only relevant in how much velocity the bullet accrues.

Example 1

Let's say your first bullet gets accelerated at $5 {m \over s^2}$ for some time. Velocity is the integral of acceleration, which in this case is $V(t)=A*t$$\leftrightarrow t={V \over A}. Plug the numbers in, we get t={100 {m \over s} \over 5 {m \over s^2}}$$={100 \over 5}{m \over s}{s^2 \over m}$$=20 s. Going the other way, we know F=MA$$=100g{1 kg \over 1000g}5 {m \over s^2}$$=0.5 {kg \over m\cdot s^2}$$=0.5N$. Incidentally, position is the integral of velocity, which gives us $D={A \over 2}t^2$$=2.5 {m \over s^2}(20s)^2$$=1000m$. Quite the barrel length.

We can do the same thing for the second bullet to get $t=10s$, $F=1N$ and $D=500m$.

Using all those numbers, let's start from the beginning. The gun has some gas that pushes on the bullet. In reality, it's not a constant force, but let's approximate. So the first gun pushes with 0.5 N of force. This force accelerates the bullet along the 1000m barrel for 20 seconds, after which time it's made it to 100 ${m \over s}$.

The second gun pushes with 1 N of force, accelerating the bullet along the 500m barrel for 10 seconds, after which time it's also at 100 ${m \over s}$.

Now they both impact the targets at the same velocity with the same mass. Momentum is given by $P=MV$ (not sure where p comes from, but obviously two Ms would be confusing). Since mass and velocity are the same for both bullets, we have the same momentum, $P=0.1 kg\cdot 100 {m \over s}$$=10 {kg \over s}=10 N\cdot s. The bullets will impact with the same momentum, which will apply the same force and, assuming they're the same shape and material, they'll do the same damage. (Harder and/or denser materials will tend to penetrate deeper, doing more damage to a smaller area.) Example 2 Let's use Rachit's comment to help our second example. Let's say the first bullet is just like the first bullet from the first example. 0.5 N accelerates the 0.1 kg bullet at 10 m \over s^2 for 20 seconds down a 1000m barrel. Now, the second bullet is more complex. Because we want it to be. And math is fun. :) First, 5 N accelerates the 0.1 kg bullet at 50 m \over s^2 for 20 seconds down a 10000m barrel, at which point it's traveling 1000 {m \over s}. Now, let's say these bullets have metal jackets which happen to be magnetic. Air resistance plays an effect here, so let's say this is happening on the moon for now. The bullets both go flying through the airless "atmosphere" of the moon. But the second bullet hits a series of magnetic fields that decelerate it. Let's say the magnets put -1 N of force on the second bullet, which is -10 {m \over s^2}. It decelerates over 90 seconds. The equations get a little more complex because we have a starting velocity to deal with. D={A \over 2}t^2+V_0t and V=At+V_0. We can plug in values to get D={-10 \over 2}{m \over s^2}(90 s)^2+1000{m\over s}(90 s)$$=49500m$ of magnets, and $V=-10{m \over s^2}90 s+1000{m \over s}$$=100 {m\over s}$.

In this case, it's accelerated way up, then decelerated again, but momentum is still $P=MV$ regardless of how it got to this velocity. $P=10N\cdot s$ just like before.

We could do this on Earth, but we'd have to add in air resistance, which requires differential equations or non-analytic solutions which gets a lot more complex. But the same principle applies: momentum is all about the mass and velocity at impact. On the subject of realism, bullets normally accelerate with much greater forces over much smaller distances to much higher speeds. But the same basic math applies. Two bullets with the same mass, size, shape and velocity will have the same momentum and hit the targets just as hard.

• I understand what you're trying to say here, but again, lets consider a scenario where using whatever means we got the bullets to a constant acceleration of 5m/s2 and 10m/s2 respectively. And also, using any means we ensured that when they impact the packets, their velocity's are same, 100m/s. If we did manage to create such a scenario, would the impact on both packets be the same, or if it would be different, how different would it be? – Rachit Sep 28 '15 at 6:30
• And the bullets don't have to leave the gun at the same velocity, they have to attain the same velocity during impact. – Rachit Sep 28 '15 at 6:32
• @Rachit At and after impact, there will be no difference in the two scenarios. – Praneet Srivastava Sep 28 '15 at 6:57
• @MichaelS - thank you so much for the examples. That helped me understand. – Rachit Sep 29 '15 at 8:00