A hunter finds his target standing on top of a frozen lake. He shoots with his rifle but in doing so, he too was thrown back over the lake and injured himself. Why did this happen?
I'm thinking Newton's 3rd law but aside from that don't really know how to explain.
FIRST law is in the absence of an EXTERNAL force the momentum of bullet/hunter system remains constant
momentum before firing = 0
so
0 =mass of hunter * velocity of hunter + mass of bullet * velocity of bullet
so
v hunter = - ( mass bullet/mass hunter ) vbullet
You're correct! This situation can be explained by Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction.
When the hunter shoots the rifle, the rifle exerts a force on the bullet, propelling it forward. In response to this action, the bullet exerts an equal and opposite force on the rifle according to Newton's third law. This backward force pushes the hunter back.
Now, let's consider the hunter's position on the frozen lake. Since the hunter is standing on a surface with low friction (the ice), the force from the recoiling rifle can cause the hunter to slide backward with relatively little resistance. This is why the hunter is thrown back over the lake.
Additionally, the hunter getting injured could be a result of the sudden and unexpected force acting on their body as they are pushed backward. It's important to note that this situation highlights the importance of proper gun safety and bracing oneself when firing a powerful firearm.