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Linear Momentum

Home > Study Notes > Linear Momentum
Published in: Mathematics
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linear momentum

Elizabeth C / Kochi

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Qualification: M.Sc. & B.Ed.

Teaches: Physics, Chemistry

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  1. 6.01 Q: Two trains, Big Red and Little Blue, have the same velocity. Big Red, however, has twice the mass of Little Blue. Compare their momenta. 6.01 A: Because Big Red has twice the mass of Little Blue, Big Red must have twice the momentum of Little Blue. 6.02 Q: The magnitude of the momentum of an object is 64 kilogram-meters per second. If the velocity of the object is doubled, the magnitude of the momentum of the object will be (A) 32 kg•m/ s (B) 64 kg•m/ s (C) 128 kg•m/ s (D) 256 kg•m/ s 6.02 A: (C) if velocity is doubled, momentum is doubled. Because momentum is a vector, the direction of the momentum vector is the same as the direction of the object's velocity. 6.03 Q: An Aichi D3A bomber, with a mass of 3600 kg, departs from its aircraft carrier with a velocity of 85 m/ s due east. What is the plane's momentum? p=mv Now, assume the bomber drops its payload and has burned up most of its fuel as it continues its journey east to its destination air field. 6.04 Q: If the bomber's new mass is 3,000 kg, and due to its reduced weight the pilot increases the cruising speed to 120 m/ s, what is the bomber's new momentum? 6.04 A: 6.05 Q: Cart A has a mass of 2 kilograms and a speed of 3 meters per second. Cart B has a mass of 3 kilograms and a speed of 2 meters per second. Compared to the inertia and magnitude of momentum of cart A, cart B has (A) the same inertia and a smaller magnitude of momentum (B) the same inertia and the same magnitude of momentum (C) greater inertia and a smaller magnitude of momentum (D) greater inertia and the same magnitude of momentum 6.05 A: (D) greater inertia and the same magnitude of momentum. Assume the D3A bomber, which had a momentum of 3.6 x 105 kg•m/ s, comes to a halt on the ground. What impulse is applied? 6.06 A: Define east as the positive direction: 6.07 Q: Calculate the magnitude of the impulse applied to a 0.75-kilogram cart to change its velocity from 0.50 meter per second east to 2.00 meters per second east. 6.07 A: 6.08 Q: A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30 kilogram•meters per second, strikes an obstacle. The obstacle exerts an impulse of 10 newton•seconds to the west on the block. The speed of the block after the collision is (A) 1.7 m/ s (B) 3.3 m/ s (C) 5.0 m/ s (D) 20 m/ s
  2. 6.08 A: (B) 6.09 Q: Which two quantities can be expressed using the same units? (A)energy and force (B) impulse and force (C) momentum and energy (D) impulse and momentum 6.09 A: (D) impulse and momentum both have units of kg•m/ s. 6.10 Q: A 1000-kilogram car traveling due east at 15 meters per second is hit from behind and receives a forward impulse of 6000 newton-seconds. Determine the magnitude of the car's change in momentum due to this impulse. 6.10 A: Change in momentum is the definition of impulse, therefore the answer must be 6000 newton-seconds. 6.18 Q: A 2000-kg car traveling at 20 m/ s collides with a 1000-kg car at rest at a stop sign. If the 2000-kg car has a velocity of 6.67 m/ s after the collision, find the velocity of the 1000-kg car after the collision. 6.18 A: Call the 2000-kg car Car A, and the 1000-kg car Car B. You can then create a momentum table as shown below: Objects Car A Car B Total Momentum Before (kg•m/ s) 2000 x 20 = 40,000 1000 xo = o 40,000 Momentum After (kg•m/ s) 2000 x 6.67 = 13,340 1000 x vB = IOOOvB 13,340 + IOOOvB Because momentum is conserved in any closed system, the total momentum before the event must be equal to the total momentum after the event. 40000 = 13,340 + IOOOvB 40000-13340 1000 -26.7 m/s Not all problems are quite so simple, but problem solving steps remain consistent. 6.22 Q: A 4-kilogram rifle fires a 20-gram bullet with a velocity of 300 m/ s. Find the recoil velocity of the rifle.
  3. 6.22 A: Once again, you can use a momentum table to organize your problem-solving. To fill out the table, you must recognize that the initial momentum of the system is O, and you can consider the rifle and bullet as a single system with a mass of 4.02 kg: : Objects Bullet Total Momentum Before (kg•m/ s) 0 Momentum After (kg•m/ s) 4 x vrecoil (. 300) = 6 6+4 x vrecoil Rifle Due to conservation of momentum, you can again state that the total momentum Due to conservation of momentum, you can again state that the total momentum before must equal the total momentum after, or 0 = 4vrecoil + 6. Solving for the recoil velocity of the rifle, you find: 0=6 +4 x vrecoil Vrecoil =-6/4 = -1.5m/s

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