HW#3b Page 1 of 6 2D motion: Rule of thumb: treat horizontal directions motion separately from vertical motion. Initial velocity v0 may have two components: in x direction and y direction, v

2D motion: Rule of thumb: treat horizontal directions motion separately from vertical motion.

Initial velocity v0 may have two components: in x direction and y direction, v

or v

. Use the given angle and sin or cos to calculate v

or v

.

A few questions have only x direction initial velocity, zero launch angle: v

=v

, v

=0.

Horizontal: X direction, constant velocity. no acceleration. set x=0 at the initial point. so that x=v

*t. (v0x is initial velocity's in horizontal direction)

If you can calculate the time t, you know the corresponding x position easily.

Vertical: Y direction, constant acceleration due to gravity, (y direction motion is the same as the question we asked in the first quiz. )

set ground to be y=0, y

is equal to the y value at t=0 (platform, cliff, etc...), set upward to be the positive direction and acceleration in y direction a = -g.

If you know y direction v

, y

, a, maybe also final y position or final y velocity, you should be able to solve y direction.

Rule of thumb: treat horizontal directions motion separately from vertical motion. Write 2 separate motion equations in x or y directions.

What limits the travel time? y direction motion equation.

What limits the maximum height? y direction motion equation.

What determines the horizontal distance? initial velocity in x direction and the total time.

Another good advice is to form study groups and discuss with your peers.

After the HW is due, you can request webassign to release the keys. Within a few days, detailed solution steps will be put on my website. You should still work hard on the HW questions after the due date and understand them all. Otherwise, you will fail in Quiz or Exams.

Problem 1. A cannonball is fired out of a cannon at [50] m/s at [45] degrees above horizontal, on level ground.

(a) How long will the cannonball be in the air?

(b) What is the maximum height above the ground that it will reach?

(c) How far will the cannonball travel horizontally?

Solution. We have to make a drawing that shows the

components of the initial velocity.

   

   

0 0

0 0

cos

50 m/s cos 45 35.36 m/s sin

50 m/s sin 45 35.36 m/s

x

y

v v

v v







  



  

(They happen to be the same because the angle is exactly 45 degrees.)

(a) We will calculate the time to get to the top of the path, using v

 v

 gt , with v

= 0 and v

=35.36 m/s. This gives

x y

v0

h

R

 v0y

v0x

 

 

0

2

35.36 m/s

3.61 s 9.8 m/s

v

t  g  

The total time in the air is twice this, or 7.22 s.

(b) We will use the equation for y, with y

= 0 and v

=35.36 m/s:

     ^{ }

1 2

0 2

2 2 1

35.36 m/s 3.61 s

9.8 m/s 3.61 s 63.79 m

y  v t

 gt

 



(c) The horizontal distance traveled is just the horizontal velocity, equal to v

, times the total time in the air. x  v t

  35.36 m/s 7.22 s    255.3 m

Problem 2 (Walker Ch. 4 P. 25). A cork shoots out of a champagne bottle at an angle of [34.0]

degrees above the horizontal. If the cork travels a horizontal distance of 1.30 m in [1.35] s, what was its initial speed?

Solution. This is a pretty darn easy problem, if we assume that it lands at the same vertical position as it starts out, and assuming we remember the range equation,

2

 

0

sin 2 R v

g 



We can just solve this for the initial speed v

:

 

  ^{ }

 

0

2

sin 2

9.8 m/s 1.30 m sin 68 3.71 m/s v gR

 

 



OK. This is not the right answer. Because it was wrong to assume that the cork had landed.

And we did not use the time that was given. It is risky to use these kind of equations without carefully examine the conditions.

We should not forget the basic rules to treat 2D motions.

So . . . the information given tells us that the horizontal component of the cork's velocity, which is constant, is equal to

1.30 m

0.970 m/s 1.34 s

x

v x t

   



Now, referring to the diagram, the relation for the x component of v

gives

0

0

cos

0.970 m/s

1.170 m/s cos cos 34

x

x

v v

v v







   



And that's right.

y y

 34

v

x

v

Problem 3. In Denver, children bring their old jack-o-lanterns to the top of a tower and compete for accuracy in hitting a target on the ground (figure 4-18). Suppose that the tower height is h = [9.60] m, and that the bulls-eye's horizontal distance is d = [3.7] m from the launch point. If the pumpkin is thrown horizontally, what is the launch speed needed to hit the bulls-eye? (Neglect air resistance.)

Figure 4-18

Solution. The time to fall a distance h under gravity, with no initial velocity in the vertical direction, is given by

1 2

0 0 2

1 2

0

0

y y v t

gt

h gt

  

    giving

 

 

1 2 2

2

2 9.60 m

2 1.400 s

9.8 m/s gt h

t h g



  

Then the horizontal velocity, which is constant, is given by

 

 

0

3.7 m

2.643 m/s 1.400 s

x x

v v x t

    



Problem 4: Walker P. 4.10. An astronaut on the planet Zircon tosses a rock horizontally with a speed of [6.35] m/s. The rock falls through a vertical distance of 1.40 m and lands a horizontal distance of 7.75] m from the astronaut. What is the acceleration of gravity on Zircon? (Neglect air resistance.)

It is better to solve this problem all in symbols, putting in numerical values later.

h g

d v0x

y

x

First - find the time for the vertical motion to take place. This is falling from y = h to y = 0, with v

= 0.

1 2

0 0 2

1 2 2 2

0 (0)

2 2

y y

y y v t gt

h gt

gt h

t h g

  

  





Next, analyze the horizontal motion. We know the initial velocity in the x direction, so the distance is

0 0

2

x x

d v t v h

  g

Finally, what were we supposed to find? The value of g! So, solve the equation above for g.

   

 

2

2 0

2 2 0 2

2 2

2

2 6.35 m/s 1.40 m

2 1.880 m/s

7.75 m

x

x

d v h g g v h

d



  

Problem 5: A small car collides with a large truck.

(a) Is the force experienced by the car greater than, less than, or equal to the force experienced by the truck?

greater than less than

equal to Explain.

Key: These forces form an action-reaction pair, and hence the force experienced by the car has the same magnitude as the force experienced by the truck. The forces point in opposite directions, however.

(b) Is the acceleration experienced by the car greater than, less than, or equal to the acceleration experienced by the truck?

greater than less than equal to Explain.

Key: The car, with its smaller mass, experiences the larger acceleration.

Problem 6: A skateboarder on a ramp is accelerated by a nonzero net force. For each of the following statements, state whether it is always true, never true, or sometimes true.

(a) The skateboarder is moving in the direction of the net force.

always true never true

sometimes true

(b) The acceleration of the skateboarder is at right angles to the net force.

always true never true sometimes true

(c) The acceleration of the skateboarder is in the same direction as the net force.

always true never true sometimes true

(d) The skateboarder is instantaneously at rest.

always true never true

sometimes true

The ramp can be any shape or even a circle. All we know about the object is that the total force on it is not zero. The question is asking based on the fact that the total force is always not zero, what can we know about its motion:

Will object always move on the same direction as the net force on it? v same direction as F, never or sometimes or always?

Will The acceleration of the object be at right angles to the net force? a is perpendicular to F, never or sometimes or always?

The acceleration of object is in the same direction as the net force. never or sometimes or always?

The object is instantaneously at rest. never or sometimes or always? Recall the concept of instantaneous speed or instantaneous velocity.

instantaneously at rest means v=0 for that particular time and only at that particular time. For example, when you throw something upward, when it reaches the top, its v-0, but only for that moment. v=0 doesn't last for a particular amount of time. v keeps changing.

Solution or Explanation

(a) This is sometimes true. For example, if the skateboarder is moving down the ramp the motion is in the same direction as the net force. On the other hand, if the skateboarder moves up the ramp, the direction of motion is opposite to the direction of the net force. In general, there is no reason for the direction of motion to be in the same direction as the net force. (b) This is never true. The acceleration of an object is always in the direction of the net force. (c) This is always true, as mentioned in part (b).

(d) This is sometimes true. Again, the fact that an object accelerates in a certain direction tells you nothing about its direction of motion, or whether it is instantaneously at rest. An example would be a skateboarder coasting upward on a ramp. At the skateboarder's highest point, he or she is

instantaneously at rest, though still accelerating in a direction pointing down the ramp.

Problem 7: You jump out of an airplane and open your parachute after a brief period of free fall. To decelerate your fall, must the force exerted on you by the parachute be greater than, less than, or equal to your weight?

greater than less than equal to

Solution or Explanation

To cause acceleration, the upward force exerted on you by the parachute must be greater in magnitude than your weight. In this case, the net force acting on you is upward, as is your acceleration. Since your acceleration and velocity are in different directions, you will decelerate. See Figure 2-11 for further details.

Problem 8. You hold a brick at rest in your hand.

(a) How many forces act on the brick?

(b) Identify where these forces come from.

(c) Are these forces equal and opposite?

(d) Are these forces an action-reaction pair?

Solution. There are two forces acting on the brick: its weight, acting downwards; and a force (N) acting upwards on the brick, exerted by the hand.

The forces are equal and opposite, adding up to zero, since the brick is not accelerating.

And - they certainly are not an action-reaction pair. The two forces of an action-reaction pair act on different bodies.

An action-reaction pair is between two object. The force that A applies to B and the force B applies to A are an Action-reaction pair.

The normal force from the hand to the brick and the pressing force from the brick to the hand are an action-reaction pair. They are always equal and opposite.

Problem 9. Referring to the previous question, you are now accelerating the brick upward.

(a) How many forces act on the brick?

(b) Identify where these forces come from.

(c) Are these forces equal and opposite?

(d) Are these forces an action-reaction pair?

Solution. Now the normal force is larger than the weight. There are still two forces, from the same two sources; but they are not equal, nor are they an action- reaction pair.

The normal force from the hand to the brick and the pressing force from the brick to the hand are an action- reaction pair. They are always equal and opposite.

N and mg are two forces applied to one object (the brick). N and mg can be different in size.

W=mg N

W=mg N