Question #95922

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

Expert's answer



a) From the conservation of momentum, 


m1v1+m2v2=m1v1′+m2v2′m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'

v2′=m1m2(v1−v1′)+v2v_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,


v2′=1.62.1(4−3)−2.5=−1.74msv_2'=\frac{1.6}{2.1} (4 - 3) -2.5 =-1.74\frac{m}{s}

b)From the conservation of energy,


0.5m1v12+0.5m2v22−0.5m1v1′2−0.5m2v2′2=0.5kx20.5 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

m1(v12−v1′2)+m2(v22−v2′2)=kx2m_1 (v_1^2 - v_1'^2) + m_2 (v_2^2 - v_2'^2) = kx^2

1.6(42−32)+2.1(2.52−1.742)=600x21.6 (4^2 - 3^2) + 2.1 (2.5^2 - 1.74^2) = 600 x^2

x=0.16 mx=0.16\ m


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