Question

Problem 4: Two carts with masses 250. g and 200. g are free to slide without friction on a horizontal airtrack. The 250.g cart is pushed against a spring, compressing it by 3.00 cm, then released. It slides towards the initially stationary 200 g mass and collides with it. After the collision the 200. g mass has a forward velocity of 0.450 m/s and the 250. g mass has a forward velocity of 0.250 m/s a) What was the velocity of the 250. g cart just before the collision? b) What was the spring constant?

Answer 1

given m = 250g = 0,25kg me = 200g = 0.20 ko before collision, velocity of my is v & vele lyy my is = 0 Aftor collision, velocity fm, is 1 0. 250 ml velocity fm, is g = 0.450 m/s Jm As momentum is conserved, intel momentum . mo,+ m, o, - mv tmave mo, to = mvit ma ve [u, za o= mivit mave – 0.25 0.25+ 0.20 0.45 m O. 25 Jo = 0,61 m/s relocity of 250g cart before collision is 0.61 m/s

spring compression x23cm = 0.03 m Let the spring constant be 'k' A Energy is conserved. Kinetic energy of m, before Collision is equal to Elastic Potential energy stored in spring 1 m, ut = 1Kx² 1K K = mo2 = 0.25X (0.610² (0.03)2 K = 103.36 N.nl

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