AP physics newton's motion questions.. is my answer correct?


by r3dxP
Tags: correct, motion, newton, physics
r3dxP
#1
Sep23-05, 12:38 AM
P: n/a
1. The net force acting on an object is suddenly reduced to zero and remains zero.
As a consequence, the object

A. stops abruptly

B. stops during a short time interval

C. changes direction

D. continues at a constant velocity

E. change velocity in an unknown manner

i got C.

2. A bullet of mass 1.8 X 10-3 kg, moving at 500 m/s impacts a large fixed block of wood and travels 6 cm before coming to rest. Assuming that the acceleration of the bullet is constant, find the force exerted by the wood on the bullet

do not know

3. A traffic light is supported by 2 wires as shown below. Is the tension in the wire that is more nearly vertical greater or less than the tension in the other wire?

T1 = T2 tan 30degrees
T1 = .6T2
the wire that is more vertical(T2) has a greater thension than that of the other wire (T1)







4. Your car is stuck in a mud hole. You are alone, but you have a long, strong rope. Having studied physics, you tie the rope tautly to a telephone pole and pull on it sideways as shown below.

A. Find the force exerted by the rope on the car when the angle is 3o and you are pulling with a force of 400 N but the car does not move.

2Tsin3degrees=400N == T=3820N ~= 4000N
B. How strong must the rope be if it takes a force of 600 N to move the car when is 4o? (I know Im not much of an artist!)

2Tsin4degrees=600N == T=4301N ~= 4000N






5. A 65-kg student weighs himself on a scale mounted on a skateboard that is rolling down an incline as shown below. Assume there is no friction so that the force exerted by the incline on the skateboard is normal to the incline. What is the reading on the scale if = 30degrees?

N = mgcos30degrees where m=65kg and g=9.8m/s^2; F=552N















6. Five forces pull on the 4.0 kg box as shown below. Find the boxs acceleration as a magnitude and direction.

y forces: 5N+14sin30degrees - 17N = ma
x forces: 3N + 14cos30degrees - 11N = ma
umm from here, i didnt know what to do.. my attempt was :
sin90/6.5 = sin x/4.1 x=39.1degrees; 6.5N/4.0kg = 1.6m/s^2; due 39.1+270=309.1degrees due east.














from 7-12, i do not know how to do any of these. any help will be appreciated.
7. A block with mass m is supported by a cord C from the ceiling, and a similar cord D is attached to the bottom of the block. (See below) Explain this: If you give a sudden jerk to D, it will break, but if you pull on D steadily, C will break.



8. In a laboratory experiment, an initially stationary electron (mass = 9.11 X 10-31 kg) undergoes a constant acceleration through 1.5 cm, reaching a speed of 6.0 X 106 m/s at the end of that distance.

A. What is the magnitude of the force accelerating the electron?

B. What is the weight of the electron?



9. A golfer chips a ball toward a vertical wall 20.0 m straight ahead, trying to hit a 30.0 cm diameter red circle painted on the wall. The target is centered about a point 1.20 m above the point where the wall intersects the horizontal ground. On one try, the ball leaves the ground with a speed of 15.0 m/s and at an angle of 35.0o above the horizontal ground and then hits the wall on a vertical line through the circles center.

A. How long does the ball take to reach the wall?

B. Does the ball hit the red circle (give the distance to the circles center)?

C. What is the speed of the ball just before it hits?

D. Has the ball passed the highest point of its trajectory when it hits?



10. A baseball is hit at Fenway Park in Boston at a point 0.762 m above home plate with an initial velocity of 33.53 m/s directed 55.0o above the horizontal. The ball is observed to clear the 11.28 m high wall in left field (the Green Monster!) 5.00 s after it is hit, at a point just inside the left-field foul line pole. Find

A. the horizontal distance down the left field foul line from home plate to the wall

B. the vertical distance by which the ball clears the wall

C. the horizontal and vertical displacements of the ball with respect to home plate 0.500 s before it clears the wall.



11. In the figure below, a ball is fired at an angle of 30.0o from the horizontal, landing 3.00 s later after traveling a horizontal distance of 100 m.

A. At what height h above the firing level does the ball land?

B. At what speed is it fired?

C. At what speed does it land?






12. You are to throw a ball with a speed of 12.0 m/s at a target that is 5.00 m above the level at which you release the ball (see below). You want the balls velocity to be horizontal at the instant it reaches the target.

A. At what angle from the horizontal must you thrown the ball?

B. What is the horizontal distance from the release point to the target?

C. What is the speed of the ball just as it reaches the target?
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Pengwuino
Pengwuino is offline
#2
Sep23-05, 12:45 AM
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1. Wrong, for one, consider what logic made you believe it was C. Then note Newton's First Law. (The 2nd Law, F=ma will also help you understand what a force is)

2. Well you have the mass and its initial velocity. You can use this to find the initial momentum... plus you know the final momentum will be 0. Since Force is just the change in momentum over the change in time, you can determine the force.

3&4. I've always been worrisome about these problems so I'll let someone else point you in the right direction.

5. Looks about right.
VietDao29
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#3
Sep23-05, 09:39 AM
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#3, I can't tell because there's no picture there. But yes, the tension in the wire that's more nearly vertical greater than the tension in the other wire. You can break the 2 tension into x and y-component. And Ty1 - Ty2 = 0.
#6, I can't tell, either. There's no picture.
#8, you have the initial velocity of the electron, vi = 0 m / s. You have the distance the electron goes d = 1.5 cm = 0.015 m, you have its final velocity vf = 6.0 X 106 m/s (I am assume it's 6.0 X 106, not 6.0 X 106). And you know its acceleration is constant. So what equation should you use?
Shouldn't it be: vf2 = vi2 + 2ad?
From there, you can find the force acting on the electron.

#12, you want the ball's velocity to be horizontal at the instant it reaches the target, that means when it reaches the target, it velocity has no y-component.
You can use [tex]\Delta KE = Work[/tex], gravitational force is the only force acts on the ball. Or you can use [tex](E_p + E_k)_1 = (E_p + E_k)_2[/tex].
Note that the x-component does not change.

#9, you can break the velocity into x and y-component. The x-component makes the ball go forwards (or backwards). The x-component does not change (since there is no air resistance). So having the distance and the x-component, can you find out how long it takes the ball to reach the wall?
At the instant of the ball hitting the wall, can you find out how high the ball is? From there can you say if it hits the circle?
Since the x-component does not change, you just need to recalculate the y-component at the instant the ball hits the wall. Having the x, and y component, can you find out its speed?
What can you say if the y-component of the ball points up? Have its reached its highest point? How about when it points down?
#11 can be done the same.
Viet Dao,


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