Recent content by hairey94

My initial guess... Third dark fringe is 2nd order dark fringe and second bright fringe is 1st order bright fringe. Is this the correct guess?

Not sure how the problem set up initially as no diagram was provided in the question. Please help me to start with the solution. Your assumptions and educated guess are appreciated.

ok really appreciated for the help. I think the question just want to make the calculation simpler but instead it is poorly worded that it becomes so much complicated as we tried to assume many things at once.

But the question stated that 1 ms before stopping. Should we just assume 0 instead of 0.28?

Ok thank you for your help. This question really took me hours to complete despite the underlying concepts involved are actually quite simple.

can I just skip to use s = (1/2)(v'+0)##\Delta t##? Instead of finding the deceleration.

Is that speed can be used for the motion of the bullet inside the block? My guess was speed of bullet before collision is the initial speed and the acceleration is -g. But you point out my acceleration is wrong which I can accept that, but how about the initial speed of the bullet?

the suggested solution is more complicated than I am expected but thank you for your help. I've calculated using your equation and I got v' = 121.04 m s-1 vs the old v' = 116.95 m s-1 So what should I do from here?

then please suggest a solution for this question

If this question asked for internal energy, then it is a straightforward thing, just use the formula U = (3/2) nRT = (3/2) PV. But the question specifically asked for average translational kinetic energy which I'm stuck with the value of T to plug in into the formula as the question only give us...

well you tell me if it is right or wrong... that is only my attempt that lead to the incorrect answer.

That is the time for the bullet to stop inside the block = 1 ms not the collision time. During this 1 ms the bullet is already inside the block.

a = -g = -9.81 m s-2

I used the linear motion equation: s = ut + (1/2) at2

Linear Motion Equation to get the common velocity of the block and bullet just after collision: v2=u2+2gs, I set v=0 at max height of s = 0.004 m and g = -9.81 m s-2 I got u = 0.28 m/s Then I calculate the impulse of the block using formula J = mv - mu, where v=0 and u=0.28 m/s and I got J = 1.4...