A bullet of mass 10 g travelling horizontally with a speed of 50 7 m/s strikes a 490 g block placed on a frictionless track, as shown in the figure. The bullet becomes embedded in the block, and the combined system then travels towards a semicircular track of radius 0.2 m . At what location will the block hit the horizontal part after leaving the semicircular track?

Step-by-step solution

By conservation of linear momentum during the collision: m u = (m + M) v 0.01 50 7 = (0.01 + 0.49) v 0.5 7 = 0.5 v v = 7 m/s Let the block leave the semicircular track at an angle above the horizontal diameter. At this point, the normal reaction N = 0 . The component of weight along the radius provides the necessary centripetal force: (m + M)g = (m + M)v'^2 R v'^2 = gR = 10 0.2 = 2 By conservation of mechanical energy from the bottom of the track to this point: 1 2 (m + M)v^2 = 1 2 (m + M)v'^2 + (m + M)gR(1 + ) v^2