If someone could please give me a step-by step answer to thisquestion, it would

Question

If someone could please give me a step-by step answer to thisquestion, it would really make my day!   Thanks inadvance if you are able to help: A 2.50-g bullet, traveling at a speed of 425 m/s, strikes thewooden block of a ballistic pendulum.   The block has amass of 215 g.   (a)   Find the speed of the bullet/blockcombination immediately after the collision. (b) How high does the combination rise above its initialposition? Again, please work through the problem if you answerthis.   Thanks so much!   :-) If someone could please give me a step-by step answer to thisquestion, it would really make my day!   Thanks inadvance if you are able to help: A 2.50-g bullet, traveling at a speed of 425 m/s, strikes thewooden block of a ballistic pendulum.   The block has amass of 215 g.   (a)   Find the speed of the bullet/blockcombination immediately after the collision. (b) How high does the combination rise above its initialposition? Again, please work through the problem if you answerthis.   Thanks so much!   :-)

Explanation / Answer

Momentum = mv Conservation of momentum is the key: m1v1 + m2v2 =m3v3 The subscripts with "1" refer to the bullet, the subscripts with"2" refer to the block, and the subscripts with "3" refer to thebullet-block combination after the collision. Putting in numbers, we have this equation: 0.0025*425 + 0.215*0 = (0.0025 + 0.215)*v3 1.0625 = 0.2175*v3 v3 = 4.89 m/s To find how high the block goes, we need to find the kineticenergy. KE = (1/2)mv2 KE = (1/2)*0.2175*(4.892) KE = 2.60 J Now use the potential energy formula (PE = mgh) to find out howhigh this 2.60 J of energy can move the ballistic pendulum up: 2.6 = 0.2175 * 9.8 * h h = 1.21 meters

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