A small wooden block with mass 0.800 kg is suspended from the lower end of a lig

Question

A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1 56 m long. the block is initially at rest. A bullet with mass 0.0100 kg is fired at the block with a horizontal velocity v_0. the bullet strikes the block and becomes embedded in it. After the collision, the combined object swings on the end of the cord. When the block has risen a vertical height of 0.700 m, the tension in the cord is 4.62 N. What was the initial speed v_0 of the bullet? Express your answer with the appropriate units.

Explanation / Answer

The cord forms a triangle with the horizontal and vertical axes.
The height of the triangle is 1.56m - 0.700m = 0.86 m
Then the angle w/r/t vertical is = arccos(0.86/1.56) = 56.54º

Then component of the weight along the cord is:
Fg = mgcos = 0.81kg * 9.8m/s² * cos56.54º = 4.377 N
The remaining tension is due to centripetal force:
Fc = 0.24 N = mv²/r = 0.81kg * v² / 1.56m
v = 0.68 m/s

Now we can use CoE to find the post-impact velocity:
total E = PE + KE = 0.81kg*9.8m/s²*0.7m + ½*0.81kg*(0.68m/s)² = 5.74 J
Then post impact, E = KE = 5.74 J = ½ * 0.81kg * v²
v = 3.766 m/s

Now we can use CoM to find the pre-impact velocity: initial p = final p
0.01kg * v + 0.8kg * 0 = 0.81kg * 3.766m/s
v = 305 m/s

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