Answer :
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:
[tex]m_Agh=\frac{1}{2}m_Av_A^2[/tex]
where
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]m_A=0.5 kg[/tex] is the mass of the block
[tex]v_A[/tex] is the speed of the block A just before touching block B
Solving for the speed,
[tex]v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s[/tex]
Then, block A collides with block B. The coefficient of restitution in the collision is given by:
[tex]e=\frac{v'_B-v'_A}{v_A-v_B}[/tex]
where:
e = 0.7 is the coefficient of restitution in this case
[tex]v_B'[/tex] is the final velocity of block B
[tex]v_A'[/tex] is the final velocity of block A
[tex]v_A=3.83 m/s[/tex]
[tex]v_B=0[/tex] is the initial velocity of block B
Solving,
[tex]v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s[/tex]
Re-arranging it,
[tex]v_A'=v_B'-2.68[/tex] (1)
Also, the total momentum must be conserved, so we can write:
[tex]m_A v_A + m_B v_B = m_A v'_A + m_B v'_B[/tex]
where
[tex]m_B=2 kg[/tex]
And substituting (1) and all the other values,
[tex]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[/tex]
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
[tex]\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2[/tex]
where
k = 600 N/m is the spring constant
x is the compression of the spring
And solving for x,
[tex]x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(2)(1.30)^2}{600}}=0.075 m[/tex]
