Physics, 01.09.2019 10:30 dayi80

Ablock of mass m is initially at rest on a frictional less floor. the block, attached to a massless spring with spring constant k, is initially at its equilibrium position. an arrow with mass m and velocity v is shot into the block. the arrow sticks in the block. what is the maximum compression of the spring

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Answer from: Lujfl

Let's understand the situation carefully. once we understand it, the answer is very simple.

The arrow with mass m, velocity v strikes the block of mass M which was at rest. The block and arrow now move together with velocity V. From conservation of momentum,
$mv=(m+m)V \\ \\ V= \frac{mv}{M+m}$

Now the block and arrow will move away from the equilibrium position of spring and thus they will expand the spring. The kinetic energy of the block and arrow will decrease and the potential energy of the spring will increase. It will go on till all the KE is converted into PE of spring. After that the Spring will start to compress and the block and arrow will gain KE. At the mean position, their KE will be same as initial as there is no dissipation of energy(just after the arrow had stroke the block). Now the block will start compressing the spring and again the KE of the block and arrow would decrease gradually and the PE of spring will increase again. It will continue till the KE of block and arrow reaches 0. That is the point of maximum compression of spring and all the KE of the block and arrow is converted to PE of spring as there is no dissipation of energy.

So from conservation of energy,
KE of block and arrow after the strike = PE of spring at maximum compression point
$\frac{1}{2}(M+m)V^2= \frac{1}{2}k x^{2}$

where x = the distance between equilibrium position and maximum compression point

$\frac{1}{2}(M+m) ( \frac{mv}{M+m})^{2} = \frac{1}{2}k x^{2} \\ \\ x^{2} = \frac{ (mv)^{2} }{k(M+m)} \\ \\ x = \frac{ mv }{ \sqrt{k(M+m)}}$

So the maximum compression is $\boxed{\frac{ mv }{ \sqrt{k(M+m)}}}$

Answer from: Quest

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this. hope that you like the video and that it explains more in detail to you.

that is the link.

Answer from: Quest

snell's law

refractve index = sin incidence / sin refraction

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