A 2.8 kg block slides along a frictionless tabletop at 5.0 m/s toward a second b

Question

A 2.8 kg block slides along a frictionless tabletop at 5.0 m/s toward a second block (at rest) of mass 6.2 kg. A coil spring, which obeys Hooke's law and has spring constant k = 860 N/m, is attached to the second block in such a way that it will be compressed when struck by the moving block, see the figure.

a) What will be the maximum compression of the spring?


b) What will be the final velocities of the blocks after the collision? (Assume the initial direction of the 2.8 kg block is positive.)
i) The 2.8 kg block?

ii) The 6.2 kg block?

c) What is the kinetic energy of the system..
i) before?

ii) after?

Explanation / Answer

assuming the total energy of block 1 is used to compress the spring initial energy = .5 x m x v^2 = .5 x 2.8 x 5 x 5 = 35 .5 x k x x^2 = 35 .5 x 860 x x ^2 = 35 x = 0.2853 m intial momentum = 2.8 x 5 = 14 final momentum = m2v2 (towards right) and m1v1 (towards left) net momentum = m2v2 - m1v1 6.2v2 - 2.8v1 = 14 v2 = 2.26 + .45v1 from kinetic energy .5 x 2.8 x v1^2 + .5 x 6.2 x v2^2 = 35 2.8v1`^2 + 6.2 v2^2 = 70 putting v2 in terms of v1 2.8v1^2 + 6.2(2.26 + .45v1)^2 = 70 2.8v1 + 6.2(5.1 + 2v1 + .2v1^2) = 70 2.8v1 + 31.62 + 12.6v1 + 1.24v1^2 = 70 1.24v1^2 + 15.4v1 - 38.38 = 0 solving for v1 v1 = 2.13 m/s (towards left) v2 = 3.22 m/s (towards right) the kinetic energy remains conserved before and after = 35 J

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