Physics, 13.01.2021 08:10 williejaroid123

The 55g arrow is launched so that it hits and embeds in a 3.54kg block. The block hangs from strings. After the arrow joins the block, they swing up so that they are 0.49m higher than the block's starting point. How fast was the arrow moving before it joined the block? THE ANSWER FOR PEOPLE WITH MASTERING PHYSICS CUZ MOST OF THEM ARE WRONG ON PE = mgh = (3.54 + 0.55)kg * 9.8m/s² * 0.49m = 17.3 J

which means the post-collision KE = 17.3 J = ½ * 3.595kg * v², and

v = 3.10 m/s

Now conserve momentum: final Σ mv = initial Σ mv

3.595kg * 3.10m/s = 0.055kg * V

V = 203 m/s ≈ 200 m/s

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The 55g arrow is launched so that it hits and embeds in a 3.54kg block. The block hangs from strings...

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