A 11-g bullet with an initial speed of 306 m/s is shot directly at a 1-kg wooden
ID: 585273 • Letter: A
Question
A 11-g bullet with an initial speed of 306 m/s is shot directly at a 1-kg wooden block, which rests on a frictionless surface. What is the speed of the block, after the bullet is embedded in the block?
(Although this is not justified, calculate your answer to 4 significant figures. This will uncover any conceptual errors you might make.)
A 11-g bullet with an unknown initial speed is shot directly at a 1.2-kg wooden block, which rests on a frictionless surface. After impact, the bullet is embedded into the block, and the block moves at a speed of 2.17 m/s. What was the initial speed of the bullet?
(Although this is not justified, calculate your answer to 4 significant figures. This will uncover any conceptual errors you might make.)
A 12-g bullet with an initial speed of 301 m/s is shot directly at a 1.1-kg wooden block, which rests on a frictionless surface. The bullet passes through the block and emerges with a speed of 270 m/s. What is the speed of the block, after the bullet has passed through the block?
A 10-g bullet with an initial speed of 301 m/s is shot directly at a 1-kg wooden block, which rests on a frictionless surface. The bullet passes through the block, after which the block is moving at a speed of 0.186 m/s. What is the speed of the bullet after it has passed through the block?
A 10-g bullet with an unknown initial speed is shot directly at a 1.2-kg wooden block, which rests on a frictionless surface. The bullet passes through the block and emerges with a speed of 295 m/s, after which the block is moving at a speed of 0.178 m/s. What was the speed of the bullet before it hit the block?
Explanation / Answer
From conservation of momentum 11 * 306 = (1000+ 11) * v
v = 3.3294 m/s
Now 11 *u = 1211* 2.17
u = 238.897 m/s
For problem 3
10 * 301 = 1000* v + 10 * 207
v = 0.94 m/s
for problem 4
10 * u = 1200 *0.178 + 10 * 295
u = 316.36 m/s
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