Question 1 A 13-g bullet is fired from a 3.08-kg rifle. After the bullet leaves

Question

Question 1

A 13-g bullet is fired from a 3.08-kg rifle. After the bullet leaves the rifle, it moves at a speed of 300 m/s. What is the bullet's momentum?

QUESTION 2 A 13-g bullet is fired from a 3.81-kg rifle. After the bullet leaves the rifle, it moves at a speed of 309 m/s. What is the magnitude of the impulse imparted on the rifle?

QUESTION 3 What is the momentum of a 5.15-ounce baseball at 76 mph?

QUESTION 4 The Marlin's Jarrod Saltalamacchia catches a 66-mph fast 5.24-ounce baseball. What is the impulse imparted on his hand and glove?

QUESTION 5 A 11-g bullet with an initial speed of 293 m/s is shot directly at a 1.2-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.)

QUESTION 6 Copy of A 13-g bullet with an unknown initial speed is shot directly at a 1-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 2.45. 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.)

QUESTION 7 A 10-g bullet with an initial speed of 304 m/s is shot directly at a 1-kg wooden block, which rests on a frictionless surface. The bullet passes through the block and emerges with a speed of 270. What is the speed of the block, after the bullet has passed through the block?

QUESTION 8 A 12-g bullet with an initial speed of 299 m/s is shot directly at a 1.1-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 0.172. What is the speed of the bullet after it has passed through the block?

QUESTION 9 A 12-g bullet with an unknown initial speed is shot directly at a 1-kg wooden block, which rests on a frictionless surface. The bullet passes through the block and emerges with a speed of 300, after which the block is moving at a speed of 0.157. What was the speed of the bullet before it hit the block?

Explanation / Answer

1) Momentum = Mass * Velocity = (13 grams) * (300 m/s) = 3.9 Kg-m/s.

2) Impulse = Final Momentum - Initial Momentum,
Initial Momentum of system = 0 Kg-m/s,
Conserving momentum shows that momentum of bullet and rifle after firing will be same but in opposite directions,
Final Momentum of bullet = Final Momentum of Rifle,
=> Final Momentum of Rifle = (13 grams) * (309 m/s) = 4.017 Kg-m/s,
Impulse = 4.017-0 = 4.017 Kg-m/s.

3) 5.15 ounces = 0.146 Kg,
76 Miles per hour = 34 Meters per second.
Momentum = 0.146 * 344.964 Kg-m/s.

4)5.24 ounces = 0.148 Kgs,
66 Miles per hour = 29.5 Meters per second,
Final Velocity = 0 m/s,
Impulse = 0 - (0.148*29.5) = 4.3666 Kg-m/s.

5)Here Bullet and Block are two bodies whose momentums are to be conserved.
Let Initial velocity of bullet be 'u' m/s,
After being shot the bullet is embedded into the Block and travels at the same speed as the block.
Initila Momentum of Block = 0 Kg-m/s,
Initial Momentum of Bullet = Mass of Bullet * Initial velocity of bullet = 0.013*u Kg-m/s,
Final Momentum of Bullet = Mass of Bullet * Final Velocity of bullet = 0.013*2.45 = 0.03185 Kg-m/s,
Final Momentum of Block = Mass of Block * Final Velocity of Block = 1 * 2.45 = 2.45 Kg-m/s,
Equating Initial and Final Momentums => 0.013*u =  0.03185 + 2.45
=> u = 190.9115 m/s.
Hence Initial velocity of the bullet is 190.9115 m/s.

7)Mass of Bullet = 10 grams,
Mass of Block = 1 Kg,
Initial Velocity of Bullet = 304 m/s, => Initial Momentum of Bullet = 3.04 Kg-m/s,
Initial velocity of Block = 0 m/s, => Initial Momentum of Block = 0 Kg-m/s,
Final velocity of bullet = 270 m/s => Final Momentum of bullet = 2.70 Kg-m/s,
Let final velocty of Block be 'v' m/s,
Conserving Momentum => 3.04 + 0 = 2.70 + 1*v
=>v = 0.34 m/s.

8) Bullet Mass = 0.012 Kg,Initial Velocity of Bullet = 299 m/s, Initial Momentum = 3.588 Kg-m/s,
Block mass = 1.1 Kg, Initial Velocity of Block = 0 m/s, Initial Momentum of Block = 0 Kg-m/s,
Final velocity of Block = 0.172 m/s, Final Momentum of Block = 0.1892 Kg-m/s,
Let Final velocity of bullet be 'v' m/s, Final momentum of bullet = 0.012*v,
Conserving Momentum => Final velocity of Bullet = 283.2333 m/s.

9) Bullet Mass = 0.012 Kg, Let Initial Velocity of Bullet = 'u' m/s, Initial Momentum = 0.012*u Kg-m/s,
Block mass = 1 Kg, Initial Velocity of Block = 0 m/s, Initial Momentum of Block = 0 Kg-m/s,
Final velocity of Block = 0.157 m/s, Final Momentum of Block = 0.157 Kg-m/s,
Final velocity of bullet = 300 m/s, Final momentum of bullet = 0.012*300 = 3.6 Kg-m/s,
Conserving Momentum => Initial velocity of Bullet = 313.0833 m/s

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