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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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