Answer to Question #95922 in Mechanics | Relativity for sara

Question

a block of mass m1=1.60kg initially moving to the right with a speed of 4.00m/s on a frictionless horizontal track collides with a massless spring attached to a second block of mass m2=2.10kg initially moving to the left with a speed of 2.50m/s. the spring constant is 600N/m
at the instant m1 is moving to the right with a speed of 3.00m/s determine the magnitudeand the direction of the velocity of m2 and the compression x of the spring

Answer 1

a) From the conservation of momentum, 

$$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$$

$$v_2'=\frac{m_1}{m_2} (v_1 - v_1') + v_2$$

Let the direction to the right be positive, the direction to the left be negative. So,

$$v_2'=\frac{1.6}{2.1} (4 - 3) -2.5 =-1.74\frac{m}{s}$$

b)From the conservation of energy,

$$0.5 m_1 v_1^2 + 0.5m_2 v_2^2 - 0.5 m_1 v_1'^2 - 0.5 m_2v_2'^2 =0.5 kx^2$$

$$m_1 (v_1^2 - v_1'^2) + m_2 (v_2^2 - v_2'^2) = kx^2$$

$$1.6 (4^2 - 3^2) + 2.1 (2.5^2 - 1.74^2) = 600 x^2$$

$$x=0.16\ m$$