Average Force on Wall from Ice Cubes' Impact

In summary, an ice cube slides down a ski-jump track in a steady stream. It moves through a net vertical distance of y = 1.40 m and leaves the track at an angle of 40.0° above the horizontal. At the highest point of its trajectory, the cube stikes a vertical wall and rebounds with half the speed it had upon impact. The cube exerts an average force of 5.24 kg on the wall.
  • #1
SamTsui86
31
0
Small ice cubes, each of mass 5.30 g, slide down a frictionless ski-jump track in a steady stream, as shown in Figure P6.71. Starting from rest, each cube moves down through a net vertical distance of y = 1.40 m and leaves the bottom end of the track at an angle of 40.0° above the horizontal. At the highest point of its subsequent trajectory, the cube stikes a vertical wall and rebounds with half the speed it had upon impact. If 10.0 cubes strike the wall per second, what average force is exerted on the wall?

p6-71.gif


I don't even know how to begin, can anyone give me some hints?
 

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  • #2
Start by figuring out how fast an ice cube is going when it hits the wall. I think you can do that part.
 
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  • #3
ok, the ice hits the wall at Vf = 6.84 m/s Vfx = 5.24 m/s and Vfy = 4.4 m/s
 
  • #4
SamTsui86 said:
ok, the ice hits the wall at Vf = 6.84 m/s Vfx = 5.24 m/s and Vfy = 4.4 m/s
If the ice hits the wall at the highest point of its trajectory, it must be moving horizontally.
6.84m/s looks too big for the final velocity. How did you get that?
 
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  • #5
I use the conservation of energy to figure out the initial velocity as it leave the track so

KE + PE top = KE + PE final
0 + mgh = 1/2mv^2 + 0
v= 5.24 m/s

then since the angle is 40 degree i did 5.24/cos40 to get 6.84 m/s
 
  • #6
SamTsui86 said:
I use the conservation of energy to figure out the initial velocity as it leave the track so

KE + PE top = KE + PE final
0 + mgh = 1/2mv^2 + 0
v= 5.24 m/s

then since the angle is 40 degree i did 5.24/cos40 to get 6.84 m/s
The first part is good. That is the total velocity. The horizontal component will be less than the total, and the vertical component will be less than the total. You have the right trig term for the problem, but in the wrong place. Fix that and find the velocity at the highest point of the trajectory.
 

Related to Average Force on Wall from Ice Cubes' Impact

1. What is the average force on a wall from the impact of ice cubes?

The average force on a wall from the impact of ice cubes can vary depending on factors such as the size and speed of the ice cubes, as well as the material and strength of the wall. However, on average, it can range from a few newtons to a few hundred newtons.

2. How does the average force on a wall from ice cubes compare to other types of impacts?

The average force on a wall from ice cubes is relatively low compared to other types of impacts, such as a person hitting the wall or a heavy object falling on it. This is because ice cubes have a lower mass and velocity, resulting in a lower force upon impact.

3. Can the average force on a wall from ice cubes cause damage?

In most cases, the average force on a wall from ice cubes is not strong enough to cause significant damage. However, if the wall is made of weaker materials or if the ice cubes are moving at high speeds, it is possible for the impact to cause cracks or dents.

4. How can the average force on a wall from ice cubes be calculated?

The average force on a wall from ice cubes can be calculated using the equation F = m * v, where F is the force, m is the mass of the ice cubes, and v is the velocity of the ice cubes upon impact. However, this calculation may not be accurate as it does not take into account the angle of impact and other variables.

5. What are some ways to reduce the average force on a wall from ice cubes?

To reduce the average force on a wall from ice cubes, you can decrease the speed at which the ice cubes are thrown or dropped, or you can use materials with higher impact resistance for the wall. Additionally, placing a barrier or cushioning material between the ice cubes and the wall can also help reduce the force of impact.

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