anonymous
  • anonymous
A gun fires a 5 g bullet at v = 400ms-1. Fired at a plate (mass of 500 g), such a bullet penetrates 5mm into it, but not through. i) Determine the speed of the plate after collision. ii) Will such plate protect you from the speeding bullet? iii) What if the mass of the plate was a) lighter; b) heavier ?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Attempted Solution: i) Assumptions: Momentum is conserved \(m_1u_1 + m_2 u_2 = (m_1 +m_2) v\) Where: \(u_1\)=the initial speed of the bullet; \(v\) = speed of plate with the bullet inside \[v=\frac{m_1 u_1}{m_1 +m_2}=4ms^{-1}\] So I will get hit by a 0.5 kg travelling at 4 ms^-1 Part need help: ii) Will I be alright if I hit by such thing? Firstly I need to know the force of the plate exerted on me so I can get an idea on how equivalently heavy it would be. \[F=ma\] But I don't know how to find the acceleration of the block.
anonymous
  • anonymous
To my knowledge, there's no way to find the force the plate exert on your body. Unless you know the contact time. If you do, we can find average force with: \[I=\int F dt\\ \Delta p=F\Delta t\] where F is average force and \(\Delta t\) is contact time of plate with your body. If you want to get an idea about the effect of moving plate with your body, I think you can find it's kinetic energy.
anonymous
  • anonymous
1.v=m1u1/(m1+m2)=3.96m/s 2.will protect as it penetrated only 5 mm,not ''pierced'' 3. a.lighter-will not protect as bullet may pierce through more , 3.b.heavier-more possibility to protect,even 5mm may not go in

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anonymous
  • anonymous
@imron07 well, if I assume the bullet decelerates constantly during collision, I could find the \(t \) of collision between the bullet and the plate. Is this \(t\) also the same duration when the plate collide with me?
anonymous
  • anonymous
@twitter : The time is different, since the plate still can move even the bullet inside already stopped (relative to plate). I think @rosemuricken has good answer to this problem..
anonymous
  • anonymous
I don't like it. Unscientific. He said it will protect just because the bullet doesn't come through. Wrong. If the force generated big enough, you can have broken ribs. Talking about force, is the force generated by the bullet equal the force generated to me by the plate?
anonymous
  • anonymous
You can think of it this way: If you push Earth with force F, would Earth push another person at it's other side with the same force?
anonymous
  • anonymous
Good point. Now about conservation of momentum. Do you think the mometum here is approximately conserved? Why? Is there a way of telling when assumption of momentum conservation is unreasonable?
anonymous
  • anonymous
Yes, it IS conserved. The momentum is always conserved if there's no net external forces on the system.\[\Sigma F_{ext}=0\\ \frac{dp}{dt}=0\\p=constant \]
anonymous
  • anonymous
During the collision, the plate move about 0.1mm. If it move lesser or further than that is the momentum still conserved?
anonymous
  • anonymous
Is there any relation between the two?
anonymous
  • anonymous
Certainly I am the external force?
anonymous
  • anonymous
I think you must specify the system first. If the system is bullet and plate only, and these two aren't in touch with another object. The momenntum is conserved. If it is in touch with your body, then an external normal force act on the bullet-plate system. Momentum is NOT conserved.
anonymous
  • anonymous
It moves 0.1mm. Please address that
anonymous
  • anonymous
It moves 0.1 mm, it means an external force made it stop. Momentum is not conserved.
anonymous
  • anonymous
And yes, you're the external force. :)
anonymous
  • anonymous
I was asking is it approximately conserved.
anonymous
  • anonymous
Most real life problem have not PRECISE momentum conservation. As a physicist is it our job to think whether we can neglect this or that variable. Nothing is exact.
anonymous
  • anonymous
Not even approximatey. It will be conserved if you include you and earth in the system.
anonymous
  • anonymous
So how much can it more so you can consider it would be acceptable to use that assumption?
anonymous
  • anonymous
I don't know the answer to that question. But if you collide two billiard ball, say, the linear momentum is approximately conserved. Even if external forces act on the two-ball system: table friction, air friction.
anonymous
  • anonymous
Now you're just taking a piss mate. If there is a collision no matter what, the time during collision is not zero. Maybe t will be of magnitude 10^-6 but it is not zero. Hence the thing must have been moving during this time.

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