Saved Bookmarks
| 1. |
Figure 10-61 is an overhead view of a rod of length 1.0 m and mass 1.0 kg that is lying stationary on a frictionless surface when three bullets hit it simultaneously. The bullets move along paths that are in the plane of the rod and per-pendicular to the rod. Bullet 1 has mass 10 g and speed 2.0 m/s. Bullet 2 has mass 20 g and speed 3.0 m/s. Bullet 3 has mass 30 g and speed 5.0 m/s. The labelled distance are alpha=10cm, b = 60 cm, and c = 80 cm. As a result of the impacts, the rod-bullets system rotates around its center of mass while the center of mass moves in a straight line over the frictionless surface. (a) What is the linear speed of the system's center of mass? (b) What is the distance between the rod's center and the system's center of mass ? ( c) What is the rotational inertia of the system about the system's center of mass? |
| Answer» Solution :(a) `6.6xx10^(-2)m//s` (B) `~~1.9xx10^(-3)m`, ( C) `8.8xx10^(-2)KG*m^(2)` | |