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For a Q in which we are firing a bullet of 0.04kg onto a wooden block which is stopped after a distance of 60cm after penetrating inside it. It’s u=90m/s.

For this Q , we can find a=-6750m/s^2. Using$ v^2 = u^2 +2as$

Now, this a can also be compared with a body moving with$ -6750 m/s^2 $.Not necessarily a wooden block needs to be there.

Then , second point. My book provides average resistive force by the block on bullet = $0.04*6750=270N.$

Now , I am confused since the mass of block is not given . How can find the average resistive force exerted by it.

One way I though of is this but I am not sure if it correct or wrong.

So , using Newton’s third law here.

  1. Force by the bullet on block = -( Force by block on bullet). OR Force by bullet on block = -(force by bullet on block ). Both are not similar here, why is that?

  2. $0.04*(-6750) $= -(mass of wooden block * -6750). So , in the end we force has to be equal. Therefore , we considered F=270N since finding force by bullet on block is not what Q asked. It asked is the average resistive force by block on bullet.

  3. Is writing average here really necessary ?

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There is no need to know the mass of the block. If it were movable yes, but in this case you can assume that is a fixed wall.

So the whole idea is that you are assuming a constant (average) deceleration of the bullet. Since you already know the deceleration on the bullet, then from the equation

$$\sum F = m\cdot a$$

you know that the only force on the X axis is the deceleration force (most likely through friction).

With that you obtain $F = 270 [N]$

Another way/route would be to calculate the kinetic energy of the bullet, and then (Eventually) equate that with the work of the friction on the system until the bullet stops.

So you would get

$$\frac{1}{2} m u^2 = F\cdot s$$ $$ F=\frac{1}{2s} m u^2 $$

Which (surprise surpise) yields $$ F=270 [N]$$

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  • $\begingroup$ Why take acc in + terms value ? $\endgroup$
    – S.M.T
    Apr 26 at 4:18
  • $\begingroup$ Q2 did I use the Newton’s third law right ? Part 1. $\endgroup$
    – S.M.T
    Apr 26 at 4:21
  • $\begingroup$ Also , is there any need to write average force ? $\endgroup$
    – S.M.T
    Apr 26 at 4:21
  • $\begingroup$ I'm not sure what you mean in first comment. About the second yes you used it correct. The final one, about the average force, Yes it is necessary. Because this assumes a constant force throughout the motion ( this is almost never the case in real experiments because of a number of reason). $\endgroup$
    – NMech
    Apr 26 at 4:39
  • $\begingroup$ When we put -6750 as acceleration in Force = m*a. Then , if I put -ve acc value. Force is also -ve right. So , is it like we then change the direction of force and make it +ve F or change direction of acceleration first or sth like this ? $\endgroup$
    – S.M.T
    Apr 26 at 5:21

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