Projectile Motion Review Homework Help

In summary, the block of mass 4.0 kg is on a frictionless tabletop and is pushed towards a wall with a spring of negligible mass and spring constant 650 N/m attached to it. After being released, the block falls 0.80 m vertically and strikes a target on the floor 1.2 m away from the edge of the table. The time elapsed from the block leaving the table to hitting the floor can be calculated using the equation s = v0t + 1/2at^2. The speed of the block as it leaves the table can be found using v = v0 + at. The distance the spring was compressed can be determined using the work energy theorem.
  • #1
kayjaydee
5
0

Homework Statement



Block I of mass 4.0 kg is on a horizontal, frictionless tabletop and is placed against a spring of negligible mass and spring
constant 650 N/m. The other end of the spring is attached to a wall. The block is pushed toward the wall until the spring has been compressed a distance x, as shown above. The block is released and follows the trajectory shown, falling 0.80 m vertically and striking a target on the floor that is a horizontal distance of 1.2 m from the edge of the table. Air resistance is negligible.
a.Calculate the time elapsed from the instant block I leaves the table to the instant it strikes the floor.

b.Calculate the speed of the block as it leaves the table.

c.Calculate the distance the spring was compressed.

Homework Equations



d=vot+1/2at^2
v=vo+at
v^2=vo^2+2ad

The Attempt at a Solution



I really forgot how to do this. We are so far ahead of this unit, now, that I lost all my projectile motion mojo.
I'm guessing you solve for time using the spring constant and one of the above equations, but that's all that I know. I'm really rusty in the projectile area. :(
 
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  • #2
Since the block leaves the table with no initial downward motion you can solve the equation
[tex] s = v_0t + \frac{1}{2}at^2 [/tex]
Where [itex] s [/itex] is the distance from the table top to the floor and [itex] a [/itex] is the acceleration due to gravity.
From this time you can use the same equation again using [itex] s [/itex] as the horizontal distance moved.
For the third part try the work energy theorem :)
 
  • #3
Okay! I think I understand now! thank you!
 
  • #4
No problem, good luck! :)
 
  • #5


As a scientist, it is important to constantly review and practice concepts in order to maintain proficiency and understanding. Projectile motion involves the analysis of an object's motion in two dimensions, taking into account both horizontal and vertical components. In this scenario, we can use the equations listed above to solve for the time elapsed, speed, and distance the spring was compressed.

To calculate the time elapsed, we can use the equation d=vot+1/2at^2, where d is the vertical distance traveled (0.80 m), vo is the initial velocity (0 m/s), and a is the acceleration due to gravity (-9.8 m/s^2). Solving for t, we get t=√(2d/a)=√(2(0.80 m)/(-9.8 m/s^2))=0.4 s. This is the time it takes for the block to travel from the table to the floor.

To calculate the speed of the block as it leaves the table, we can use the equation v=vo+at, where v is the final velocity, vo is the initial velocity, and a is the acceleration due to gravity. Since the block is released from rest, vo=0 m/s. Thus, v=0+(-9.8 m/s^2)(0.4 s)=-3.92 m/s. This is the vertical component of the block's velocity.

To calculate the distance the spring was compressed, we can use the equation v^2=vo^2+2ad, where v is the final velocity (which we just calculated to be -3.92 m/s), vo is the initial velocity (again, 0 m/s), a is the acceleration due to gravity, and d is the distance the spring was compressed. Solving for d, we get d=(v^2-vo^2)/2a=(-3.92 m/s)^2/(2(-9.8 m/s^2))=0.08 m. This is the distance the spring was compressed before the block was released.

It is important to constantly review and practice concepts like projectile motion, as they are fundamental to understanding and analyzing various physical phenomena. I encourage you to continue practicing and seeking help when needed in order to improve your understanding and proficiency in this area.
 

1. What is projectile motion?

Projectile motion is a form of motion in which an object is projected or thrown into the air, and is then subject to the force of gravity. This type of motion follows a parabolic path, and is affected by factors such as initial velocity, angle of projection, and air resistance.

2. How is projectile motion different from regular motion?

Unlike regular motion, where an object moves in a straight line, projectile motion involves both horizontal and vertical motion simultaneously. This is because the object is subject to both the force of gravity and its initial velocity.

3. What are the equations used to calculate projectile motion?

The equations used to calculate projectile motion are the equations of motion: x = x0 + v0t + 1/2at2, y = y0 + v0t + 1/2at2, v = v0 + at, and v2 = v02 + 2a(x-x0).

4. How is the angle of projection related to the maximum height and range of a projectile?

The angle of projection affects the maximum height and range of a projectile by determining the vertical and horizontal components of its initial velocity. The maximum height is reached at a 45-degree angle, and the range is maximized at a 90-degree angle.

5. How can we use projectile motion in real-life situations?

Projectile motion can be used in real-life situations such as sports, where objects such as balls, frisbees, and arrows are thrown or launched. It is also used in engineering and physics to calculate the trajectory of objects like rockets and satellites. Additionally, understanding projectile motion can help us predict the path of falling objects and make accurate calculations for activities like hunting and target shooting.

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