Answer:
0.075 m
Explanation:
The picture of the problem is missing: find it in attachment.
At first, block A is released at a distance of
h = 0.75 m
above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:
![m_Agh=\frac{1}{2}m_Av_A^2](https://tex.z-dn.net/?f=m_Agh%3D%5Cfrac%7B1%7D%7B2%7Dm_Av_A%5E2)
where
is the acceleration due to gravity
is the mass of the block
is the speed of the block A just before touching block B
Solving for the speed,
![v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s](https://tex.z-dn.net/?f=v_A%3D%5Csqrt%7B2gh%7D%3D%5Csqrt%7B2%289.8%29%280.75%29%7D%3D3.83%20m%2Fs)
Then, block A collides with block B. The coefficient of restitution in the collision is given by:
![e=\frac{v'_B-v'_A}{v_A-v_B}](https://tex.z-dn.net/?f=e%3D%5Cfrac%7Bv%27_B-v%27_A%7D%7Bv_A-v_B%7D)
where:
e = 0.7 is the coefficient of restitution in this case
is the final velocity of block B
is the final velocity of block A
![v_A=3.83 m/s](https://tex.z-dn.net/?f=v_A%3D3.83%20m%2Fs)
is the initial velocity of block B
Solving,
Re-arranging it,
(1)
Also, the total momentum must be conserved, so we can write:
![m_A v_A + m_B v_B = m_A v'_A + m_B v'_B](https://tex.z-dn.net/?f=m_A%20v_A%20%2B%20m_B%20v_B%20%3D%20m_A%20v%27_A%20%2B%20m_B%20v%27_B)
where
![m_B=2 kg](https://tex.z-dn.net/?f=m_B%3D2%20kg)
And substituting (1) and all the other values,
![m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s](https://tex.z-dn.net/?f=m_A%20v_A%20%3D%20m_A%20%28v_B%27-2.68%29%20%2B%20m_B%20v_B%27%5C%5Cv_B%27%20%3D%20%5Cfrac%7Bm_A%20v_A%20%2B2.68%20m_A%7D%7Bm_A%20%2B%20m_B%7D%3D1.30%20m%2Fs)
This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to
![\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dm_B%20v_B%27%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
where
k = 600 N/m is the spring constant
x is the compression of the spring
And solving for x,
![x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m](https://tex.z-dn.net/?f=x%3D%5Csqrt%7B%5Cfrac%7Bmv%5E2%7D%7Bk%7D%7D%3D%5Csqrt%7B%5Cfrac%7B%282%29%281.30%29%5E2%7D%7B600%7D%7D%3D0.075%20m)