Regents Physics Momentum- SHS Community Service

A 1.0 kilogram mass changes speed from 2.0 m/s to 5.0 m/s. The change in the object's momentum is (1) 3.0 kg*m/s (2) 21 kg*m/s (3) 29 kg*m/s (4) 9.0 kg*m/s

Answer: (1) 3.0 kg*m/s Explanation: The givens are the mass (1 kg), the initial velocity (2 m/s) and the final velocity (5 m/s), and we are asked to find the change in momentum, or impulse. Based on the givens, it makes the most sense to use the equation ∆p=m∆v. When the correct values are plugged in for their variables, a value of 3.0 kg*m/s is obtained

Two rocks weighing 5 newtons and 10 newtons, respectively, fall freely from rest near the Earth's surface. After 3 seconds of free fall, compared to the 5 newton rock, the 10 newton rock has greater... (1) acceleration (2) momentum (3) height (4) speed

Answer: (2) momentum Explanation: Momentum is dependent upon mass. If the 10 newton rock has a greater weight, it will also have a greater mass, and therefore, a greater momentum.

The magnitude of the momentum of an object is 64.0 kilogram-meter per second. If the velocity of the object is doubled, the magnitude of the momentum will be (1) 32.0 kg-m/s (2) 64.0 kg-m/s (3) 128 kg-m/s (4) 256 kg-m/s

Answer: (3) 128 kg-m/s Explanation: This can be solved using proportional reasoning. The equation you should use to solve is p=mv. The only thing that is changing is the velocity, which is doubling, so that will be represented by a 2. The equation will now look like p= (1)(2). This means that the momentum will double as well. 64.0 kg-m/s doubled is 128 kg-m/s.

A student drops two eggs of equal mass simultaneously from the same height. Egg A lands on the tile floor and breaks. Egg B lands intact, without bouncing, on a foam pad lying on the floor. Compared to the magnitude of the impulse on Egg A as it lands, the magnitude of the impulse on Egg B as it lands is (1) greater (2) less (3) the same

Answer: (3) the same Explanation: The masses of the eggs are equal, and they are being dropped from the same height, so their velocities will be equal as well. p=mv and J= ∆p so if the mass and velocities are the same, the momentum and impulses will be the same for both eggs as well.

In a baseball game, a batter hits a ball for a home run. Compared to the magnitude of the impulse imparted to the ball, the magnitude of the impulse imparted to the bat is (1) greater (2) less (3) the same

Answer: (3) the same Explanation: The momentum is equal and opposite to one another, so the impulse will also be equal and opposite to one another.

A 0.050 kilogram bullet is fired from a 4.0 kilogram rifle that is initially at rest. If the bullet leaves the rifle with momentum having a magnitude of 20. kilogram*meters per second, what will be the magnitude of the momentum of the rifle's recoil? (1) 0.25 kg*m/s (2) 1600 kg*m/s (3) 80. kg*m/s (4) 20. kg*m/s

Answer: (4) 20. kg*m/s Explanation: Due to the conservation of momentum, the momentum of the recoil of the rifle will be equal and opposite to that of the bullet.

A 2.0 kg rifle initially at rest fires a 0.002 kg bullet. As the bullet leaves the rifle with a velocity of 500 m/s, what is the momentum of the rifle-bullet system?

Answer: 0 kg*m/s Explanation: This kind of situation is known as an explosion, and the equation for this is 0 kg*m/s= m1v1f+m2v2f, meaning that the momentum is equal to 0.

A 7.2 kg rifle fires a 3.4 x 10-³ kg bullet at a velocity of 365 m/s north. What is the velocity of the rifle after the bullet is fired?

Answer: 0.17 m/s North Explanation: The two pieces are initially at rest together and then end up apart, so this is an explosion. This means we use the equation 0 kg*m/s= m1v1f+m2v2f. The givens are m1 (7.2 kg), m2 (3.4 x 10-³ kg) and v2f (365 m/s north), and we are solving for v1f. When the correct values are plugged in for their variables, a value of 0.17 m/s North is obtained for the velocity of the rifle.

What is the momentum of a 1200 kg car traveling at 15 m/s due east?

Answer: 1.8 x 10⁴ kg*m/s due east Explanation: The givens are the mass (1200 kg) and the velocity (15 m/s) and we are being asked to find the momentum, so we should use the equation p=mv. When the correct values are plugged in for their variables, a value of 1.8 x 10⁴ kg*m/s due east is obtained.

A 42 kg boy is riding his bike eastward when he crashes into a parked car. If the car exerts a force of 2500 N westward in 0.20 second, what was the boy's change in velocity?

Answer: 12 m/s west Explanation: The givens are the mass (42 kg), net force (2500 N west) and the time (.20 s). In order to find the change in velocity, we need to also find the change in momentum. This can be done by setting Fnet*t=m∆v, as they are both equal to the change in momentum, ∆p. When the correct values are plugged in for their variables, a value of 12 m/s west is obtained for the change in velocity.

A bomb explodes into two pieces that travel in a straight line path away from each other. The mass of the first piece is 4.6 kg and is moving at 67 m/s to the right. The mass of the second piece is 2.4 kg. What is the velocity of the second piece?

Answer: 130 m/s left Explanation: The two pieces are initially at rest together and then end up apart, so this is an explosion. This means we use the equation 0 kg*m/s= m1v1f+m2v2f. The givens are m1 (4.6 kg), v1f (67 m/s right), and m2 (2.4 kg). We are looking for the velocity of the second piece, which is moving in the opposite direction (left). When the correct values are plugged in for their variables, a value of 130 m/s left is obtained for the velocity of the second piece.

A football has a mass of 2.0 kg and a momentum of 65 kg-m/s to the left. What is the velocity of the football?

Answer: 32.5 m/s left Explanation: The givens are the mass (2.0 kg) and the momentum (65 kg-m/s left) and we are being asked for the velocity of the football. Based on the givens, it makes the most sense to use the equation p=mv. When the correct values are plugged in for their variables, a value of 32.5 m/s left is obtained. Be sure to include direction for the velocity as it is a vector quantity.

What is the magnitude of the velocity of a 25 kilogram mass that is moving with a momentum of 100. kg*m/s?

Answer: 4 m/s Explanation: The givens are the mass (25 kg) and the momentum (100. kg*m/s) and we are looking for the velocity, so we should use the equation p=mv. When the correct values are plugged in for their variables, a value of 4 m/s is obtained.

A force of 100. N acts for 0.40 seconds on a 2.0 kg soccer ball. What is the impulse given to the ball?

Answer: 40 N*s right Explanation: Impulse is defied as the change in momentum, so that is what we are looking for. The givens are the mass (2.0 kg), the applied force (100. N), and the time (0.40 seconds). The equation ∆p= Fnet*t will allow us to find the change in momentum, which is equal to the impulse. When the correct values are plugged in for their variables, a value of 40 N*s right is obtained. Be sure to include direction as impulse is a vector quantity.

A 3000. kg truck moving with a velocity of 10. m/s hits a 1000. kg parked car. The impact causes the 1000. kg car to be set in motion at 15 m/s. Assuming that momentum is conserved during the collision, determine the velocity of the truck after the collision.

Answer: 5.0 m/s right Explanation: This uses the idea of the conservation of momentum, which allows us to conclude that the total momentum before the collision is equal to the total momentum after the collision. Also, being that these trucks start out separate and are separate after the collision means that we are using the equation p1i+p2i=p1f+p2f, where p represents the momentum of each object. This can be expanded to m1v1i+m2v2i=m1v1f+m2v2f. The givens are m1 (3000 kg), m2 (1000 kg), v1i (10 m/s) and v2i (0 m/s). Because v2i is 0, that whole term with it is 0. When the correct values are plugged in for their variables, a value of 5.0 m/s right is obtained for the velocity of the truck after the collision.

A 7.3 kg duck is flying around the pond. It has a momentum of 37.7 kg*m/s. What is the duck's speed?

Answer: 5.2 m/s Explanation: The givens are the mass (7.3 kg) and the momentum (37.7 kg*m/s). We are asked for the duck's speed. Based on this information, it makes the most sense to use the equation p=mv to find the speed. When the correct values are plugged in for their variables, a value of 5.2 m/s is obtained for the speed.

A 6.1 kilogram object is moving at 8.2 meters per second east. What is the impulse needed to bring the object to rest?

Answer: 50 N*s West Explanation: The givens are the mass (6.1 kg), the initial velocity (8.2 m/s), and the final velocity (0 m/s because it is being brought to rest). We want to find the impulse needed to do this. We can find this using the equation ∆p= m(vf-vi). When the correct values are plugged in for their variables, a value of 50 N*s West is obtained for the impulse. It is in the opposite direction as it is a resisting force causing it to slow down and stop.

Miners drill a hole into a large rock in order to blow it apart. A 550.-kg portion of the rock moves off with a speed of 27 m/s. What is the velocity of the smaller portion of the rock (300 kg) after the bomb explodes?

Answer: 50 m/s left Explanation: The givens are m1 (550 kg), m2 (300 kg), and v1f (2.7m/s), and we are looking for the final velocity of the smaller rock v2f. Being that the objects start out together at rest and end up apart, this can be classified as an explosion and the equation we can use to solve this is 0 kg*m/s= m1v1f+m2v2f. When the correct values are plugged in for their variables, a value of 50 m/s left is obtained.

A mass moving with a momentum of 47.2 kgm/s receives an impulse of 28.6 N*s in the direction of motion. Calculate the final momentum of the mass

Answer: 75.8 kg-m/s Explanation: The givens are the momentum (47.2 kgm/s) and the impulse (28.6 Ns) and we want to find the total momentum of the mass. We can use the idea that impulse is equal to the change in momentum, which is represented by ∆p= pf-pi. When the correct values are plugged in for their variables, a value of 75.8 kg-m/s is obtained for the final momentum.

An unbalanced force of 4.7 Newtons acts on a 3.6 kg object for 7.0 seconds. What is the magnitude of the object's change in velocity?

Answer: 9.1 m/s Explanation: The givens are Fnet (4.7 N), the mass (3.6 kg), and the time (7.0 s). In order to find the change in velocity, we can set Fnet*t=m∆v, as they are both equal to impulse. When the correct values are plugged in for their variables, a value of 9.1 m/s is obtained for the change in velocity.

Milton Vermouth has a momentum of 110 kg-m/s and a velocity of 30. m/s north. What is the direction of the momentum?

Answer: North Explanation: The direction of the velocity is north, which means that the momentum must be directed in the same direction as well, as that is the direction that Milton is traveling in.

Which of the following is an acceptable unit for impulse?

Answer: kg*m/s Explanation: Impulse is equal to the change in momentum, and the units for that are kg*m/s, therefore, impulse must have the same units.

The momentum of the object is the product of its...

Answer: mass and velocity Explanation: The equation to find momentum is p=mv, which is the product of mass and velocity

Determine the momentum of a 10. kg pumpkin moving at 8 m/s to the east.

Answer: p= 80. kg * m/s East Explanation: The givens are the mass (10 kg) and the velocity (8 m/s east) and we are being asked to find the momentum, p. Based on the givens, it makes the most sense to use the equation p=mv. When the correct values are plugged in for their variables, a value of 80. kg * m/s East is obtained.

If the direction of the momentum of an object is west, the direction of the velocity of the object is...

Answer: west Explanation: If the momentum is west, then the velocity must also be west, as the object is moving in that direction