1. The bullet remains in the block, and after the impact the block lands d = 2.0
ID: 2026658 • Letter: 1
Question
1. The bullet remains in the block, and after the impact the block lands d = 2.02 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table?
2. What is the initial horizontal velocity of the block as it flies off the table? (assume this to be in the positive direction)
3. What is the the initial speed of the bullet?
An 6.44g bullet is fired into a 205g block that is initially at rest at the edge of a frictionless table of height h = 1.15 m (see the figure above).Explanation / Answer
1. the time to reach the floor is only dependent on its height. so its D=Vi(t)+.5(a)(t)^2. and we only care about vertical acceleration and velocity. so you get 1.15=0+4.9t^2. divide both sides by 4.9 so you have .23469=t^2. take the square root of both sides and you get .484452=t 2. if it traveled 2.02 meters in .484452 seconds then use v=d/t. so its 2.02/.484452=4.169m/s. 3. now that you have the velocity which is 4.169m/s and you have the total mass you can use the conservation of momentum to find the initial velocity of the bullet. you can stay in grams provided that you're consistent throughout the problem. the final momentum of the system is m*v which is (6.44+205)*(4.169)=881.5. now since the initial velocity of the block was zero the only thing that has initial momentum is the bullet. so we know that momentum=m*v and the momentum is 881.5=m*v and the mass of the bullet is 6.44 so we have 881.5=(6.44)(v). divide both sides by 6.44 and you get the initial velocity of 136.9m/s
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