(D) Based on Ft = m v, doubling the mass would require twice the time for same momentum change

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1 1. A car of mass m, traveling at speed v, stops in time t when maximum braking force is applied. Assuming the braking force is independent of mass, what time would be required to stop a car of mass m traveling at speed v? (A) ½ t (B) t (C) t (D) t (D) Based on Ft = m v, doubling the mass would require twice the time for same momentum change. A block of mass M is initially at rest on a frictionless floor. The block, attached to a massless spring with spring constant k, is initially at its equilibrium position. An arrow with mass m and velocity v is shot into the block. The arrow sticks in the block. What is the maximum compression of the spring? (D) Two step problem. I) find velocity after collision with arrow. II) now use energy conservation. K i = U sp(f) m av ai = (m a+m b) v f v f = mv / (m+m) ½ (m+m)v f = ½ k x, sub in v f from I 3. Two objects, P and Q, have the same momentum. Q can have more kinetic energy than P if it has: (A) More mass than P (B) The same mass as P (C) More speed than P (D) The same speed at P (C) Since the momentum is the same, that means the quantity m 1v 1 = m v. This means that the mass and velocity change proportionally to each other so if you double m 1 you would have to double m or v on the other side as well to maintain the same momentum. Now we consider the energy formula KE= ½ mv since the v is squared, it is the more important term to increase in order to make more energy. So if you double the mass of 1, then double the velocity of, you have the same momentum but the velocity of when squared will make a greater energy, hence we want more velocity in object to have more energy. 4. Multiple Correct: Consider two laboratory carts of different masses but identical kinetic energie. Which of the following statements must be correct? Select two answers. (A) The one with the greatest mass has the greatest momentum (B) The same impulse was required to accelerate each cart from rest (C) Both can do the same amount of work as they come to a stop (D) The same amount of force was required to accelerate each cart from rest (A) (C) First of all, if the kinetic energies are the same, then when brought to rest, the non conservative work done on each would have to be the same based on work-energy principle. Also, since both have the same kinetic energies we have ½ m 1v 1 = ½ m v since the velocity is squared an increase in mass would need a proportionally smaller decrease in velocity to keep the terms the same and thus make the quantity mv be higher for the larger mass. This can be seen through example: If mass m 1 was double mass m its velocity would be v / times in comparison to mass m s velocity. So you get double the mass but less than half of the velocity which makes a larger mv term 5. A mass m has speed v. It then collides with a stationary object of mass m. If both objects stick together in a perfectly inelastic collision, what is the final speed of the newly formed object? (A) v / 3 (B) v / (C) v / 3 (D) 3v / (A) Explosion. p before = 0 = p after 0 = m 1v 1f + m v f 0 = (50)(v 1f) + ()(10) 6. A 50 kg skater at rest on a frictionless rink throws a kg ball, giving the ball a velocity of 10 m/s. Which statement describes the skater s subsequent motion? (A) 0.4 m/s in the same direction as the ball's motion

2 (B) 0.4 m/s in the opposite direction of the ball's motion. (C) m/s in the same direction as the ball's motion. (D) m/s in the opposite direction of the ball's motion. (B) Explosion, momentum before is zero and after must also be zero. To have equal momentum the heavier student must have a much smaller velocity and since that smaller velocity is squared 7. Two pucks are firmly attached by a stretched spring and are initially held at rest on a frictionless surface, as shown above. The pucks are then released simultaneously. If puck I has three times the mass of puck II, which of the following quantities is the same for both pucks as the spring pulls the two pucks toward each other? (A) Speed (B) Magnitude of acceleration (C) Kinetic energy (D) Magnitude of momentum (D) In a perfect inelastic collision with one of the objects at rest, the speed after will always be less no matter what the masses. The increase of mass in mv is offset by a decrease in velocity 8. Which of the following is true when an object of mass m moving on a horizontal frictionless surface hits and sticks to an object of mass M > m, which is initially at rest on the surface? (A) The collision is elastic. (B) The momentum of the objects that are stuck together has a smaller magnitude than the initial momentum of the less-massive object. (C) The speed of the objects that are stuck together will be less than the initial speed of the less massive object. (D) The direction of motion of the objects that are stuck together depends on whether the hit is a head-on collision. (C) Since the total momentum before and after is zero, momentum conservation is not violated, however the objects gain energy in the collision which is not possible unless there was some energy input which could come in the form of inputting stored potential energy in some way. 9. An object of mass M travels along a horizontal air track at a constant speed v and collides elastically with an object of identical mass that is initially at rest on the track. Which of the following statements is true for the two objects after the impact? (A) The total momentum is Mv and the total kinetic energy is ½ Mv (B) The total momentum is Mv and the total kinetic energy is less than ½ Mv (C) The total momentum is less than Mv and the total kinetic energy is ½ Mv (D) The momentum of each object is ½ Mv (A) 10. A kg object initially moving with a constant velocity is subjected to a force of magnitude F in the direction of motion. A graph of F as a function of time t is shown. What is the increase, if any, in the velocity of the object during the time the force is applied? (A) 0 m/s (B) 3.0 m/s (C) 4.0 m/s (D) 6.0 m/s (B)

3 1976B. A bullet of mass m and velocity v o is fired toward a block of mass 4m. The block is initially at rest on a frictionless horizontal surface. The bullet penetrates the block and emerges with a velocity of (a) Determine the final speed of the block. (b) Determine the loss in kinetic energy of the bullet. (c) Determine the gain in the kinetic energy of the block. a) Apply momentum conservation. p before = p after mv o = (m)(v o/3) + (4m)(v f) v f = v o / 6 b) KE f KE i = ½ mv o ½ m (v o / 3) = 4/9 mv o c) KE = ½ (4m)(v o / 6) = 1/18 mv o

4 1981B. A massless spring is between a 1kilogram mass and a 3kilogram mass as shown above, but is not attached to either mass. Both masses are on a horizontal frictionless table. In an experiment, the 1kilogram mass is held in place and the spring is compressed by pushing on the 3kilogram mass. The 3kilogram mass is then released and moves off with a speed of 10 meters per second. a. Determine the minimum work needed to compress the spring in this experiment. In a different experiment, the spring is compressed again exactly as above, but this time both masses are released simultaneously and each mass moves off separately at unknown speeds. b. Determine the final velocity of each mass relative to the cable after the masses are released. a) The work to compress the spring would be equal to the amount of spring energy it possessed after compression. After releasing the mass, energy is conserved and the spring energy totally becomes kinetic energy so the kinetic energy of the mass when leaving the spring equals the amount of work done to compress the spring W = ½ m v = ½ (3) (10) = 150 J b) Apply momentum conservation to the explosion p before = 0 = p after 0 = m 1v 1f + m v f 0 = (1)v 1f + (3)v f v 1f = 3 v f Apply energy conservation all of the spring energy is converted into the kinetic energy of the masses 150 J = K 1 + K 150 = ½ m v 1f + ½ m v f sub in above for v f 150 = ½ (1)(3v f) + ½ (3)(v f) v f = 5 m/s v 1f = 15 m/s

5 1985B1. A kilogram block initially hangs at rest at the end of two 1meter strings of negligible mass as shown on the left diagram above. A 0.003kilogram bullet, moving horizontally with a speed of 1000 meters per second, strikes the block and becomes embedded in it. After the collision, the bullet/ block combination swings upward, but does not rotate. a. Calculate the speed v of the bullet/ block combination just after the collision. b. Calculate the ratio of the initial kinetic energy of the bullet to the kinetic energy of the bullet/ block combination immediately after the collision. c. Calculate the maximum vertical height above the initial rest position reached by the bullet/block combination. a) Apply momentum conservation perfect inelastic. p before = p after m 1v 1i = (m+m)v f v f = 1.5 m/s b) KE i / KE f ½ m v 1i / ½ (m+m)vf = 667 c) Apply conservation of energy of combined masses K = U ½ (m+m)v = (m+m)gh h = 0.11 m

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