Consult Multiple-Concept Example 8 for background pertinent to this problem. A 1.66-g bullet, traveling at a speed of 483 m/s, strikes the wooden block of a ballistic pendulum, such as that in Figure 7.14. The block has a mass of 258 g. (a) Find the speed of the bullet/block combination immediately after the collision. (b) How high does the combination rise above its initial position?
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use conservation of momentum to find the speed immediately after impact
0.00166kg x 483m/s = (0.258kg+0.00166kg) V where V is the speed of the bullet/block after collision
V=3.09m/s
now use energy conservation to find the height the system reaches
initial KE = final PE
1/2 (M+m) V^2 = (M+m) g h
h = V^2/2g = (3.09m/s)^2/2g = 0.486m
0.00166kg x 483m/s = (0.258kg+0.00166kg) V where V is the speed of the bullet/block after collision
V=3.09m/s
now use energy conservation to find the height the system reaches
initial KE = final PE
1/2 (M+m) V^2 = (M+m) g h
h = V^2/2g = (3.09m/s)^2/2g = 0.486m