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Published byLizette Ainsworth Modified over 3 years ago

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In the previous question we saw how a funnel cart works when the track is horizontal with the cart moving at a constant velocity. Now suppose that the cart is attached to a string running over a pulley at the right end of the track with a weight hanging on the end, as shown in the picture below. When the cart is held at the left end of the track and released, it accelerates across the track, with the ball ejected when it gets to the same point as before. Where will the ball land? (1) The ball will fall in front of the funnel. (2) The ball will fall behind the funnel. (3) The ball will fall IN the funnel.

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In the previous two questions, we saw how a funnel cart works when the track is horizontal. For this question we will investigate what happens when the track is tilted. Now suppose that the track is tilted, as shown in the figure below, so the cart is accelerating down the incline when it gets to the point on the track at which the ball is ejected - this time perpendicular to the track. Where will the ball fall this time? (1) The ball will fall in front of the funnel. (2) The ball will fall behind the funnel. (3) The ball will fall IN the funnel.

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**1. hits the criminal regardless of the value of vo. **

Consider the situation depicted here. A gun is aimed directly at a dangerous criminal hanging from the gutter of a building. The target is well within the gun’s range, but the instant the gun is fired and the bullet moves with a speed vo, the criminal lets go and drops to the ground. What happens? The bullet 1. hits the criminal regardless of the value of vo. 2. hits the criminal only if vo is large enough. 3. misses the criminal. Answer: 1.The downward acceleration of the bullet and the criminal are identical, so the bullet will hit its target—they both “fall” the same distance.

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**Two balls are mounted as shown on the apparatus pictured below**

Two balls are mounted as shown on the apparatus pictured below. When the system is tripped the ball at the left is released from rest and falls downward; ball at the right is projected horizontally and falls to the floor in a parabolic arc. When the apparatus is tripped, sending the balls on their respective paths to the floor: (1) the ball released from rest will arrive at the floor first. (2) the ball projected horizontally will get to the floor first. (3) the two balls will get to the floor simultaneously.

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**A battleship simultaneously fires two shells at enemy ships**

A battleship simultaneously fires two shells at enemy ships. If the shells follow the parabolic trajectories shown, which ship gets hit first? 1. A 2. both at the same time 3. B 4. need more information Answer: 3. The time a projectile spends in the air is equal to twice the time it takes to fall from its maximum height. Because the shell fired at ship A reaches a higher altitude than the one aimed at B, the former takes longer to return to sea level.

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Identical balls are launched at the same time with the same velocity from the left front end of the two-track gizmo photographed below. (Because this is a physics problem, there is no friction.) A race of the balls will then ensue. The ball on the flat track clearly proceeds across the track at a constant speed. The ball on the dipped track goes for a while at that same speed, goes faster while it is in the dipped part of the track, then returns to its original speed for the final segment of the track. Note that it also travels further. What will happen? (1) The ball on the straight track will reach the end first. (2) The ball on the track with the dip will reach the end first. (3) The race will end in a tie. The speed of the ball on the dipped track is always equal to or larger than the straight track.

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Example: An archer standing on a hill that is 10 m above the center of the target. The archer fires an arrow with a speed of 120 m/s at an angle of 5o relative to the horizontal. The arrow strikes the bull's-eye. How high does the arrow go during its flight? How far away from the archer is the target? a) We need to first find the time it takes to reach the max height. We can now determine the max height.

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**t = tto max height + tfrom max height to target = 1.1 s + 1.8 s **

b) We need to find the time from the max height to the target. We then need to determine the total time the arrow in in the air. t = tto max height + tfrom max height to target = 1.1 s s = 2.9 s We can now determine the total horizontal distance covered by the arrow. Make sure to check that you are using the correct times for each part of the trip.

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