Impulse momentum theory, a ball strikes a wall dunno if i got it right?

In summary, when a ball with a mass of 0.15kg and a velocity of 5m/s strikes a wall and bounces back with a velocity of 3m/s after being in contact for 0.1 seconds, the wall exerts an average force of 12N on the ball in the opposite direction of its initial velocity. The force can be negative if the signs of the velocities are chosen in opposite directions. This means the wall exerts a force on the ball in the direction of its motion.
  • #1
girl52
7
0

Homework Statement


a ball w/ mass of .15kg and a velocity of 5m/s strikes a wall and bounces right back with v=3m/s. ball is in contact with wall for 0.1 sec, how much force did the wall exert on the ball?


Homework Equations


Ft=mvf-mvi
F=?
t=0.1
m=.15
vf=3m/s
vi=5

The Attempt at a Solution



0.1F=(.15*3)-(.15*5)
0.1F= -.30
F= -3 N

How could the force be negative?? did i do something wrong?
 
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  • #2
Velocity and momentum are vectors, which means you have to take into account the direction of the velocities (plus or minus) before and after the collision. Your answer is not correct. If the force is negative, what does that imply?
 
  • #3
PhanthomJay said:
Velocity and momentum are vectors, which means you have to take into account the direction of the velocities (plus or minus) before and after the collision. Your answer is not correct. If the force is negative, what does that imply?

OOOH okayy i don't know how i forgot that!
i just don't know which one is negative. i made the inital velocity negative, and i got 12N for the answer...is that right?

i got:
0.1F=(.15*3)+(.15*5)
F=12

when i made the final velocity negative i got 3N

im so confused
 
  • #4
12 N for the average force is correct, but what is the direction of the 12 N force? You can choose any direction as negative, as long as you choose the opposite direction as positive. If you make the final velocity negative, then the initial velocity is positive. In which case,
0.1F = .15*(-3) - (.15*5)
F = - 12 N
How do you interpret the minus sign??
 
  • #5
PhanthomJay said:
12 N for the average force is correct, but what is the direction of the 12 N force? You can choose any direction as negative, as long as you choose the opposite direction as positive. If you make the final velocity negative, then the initial velocity is positive. In which case,
0.1F = .15*(-3) - (.15*5)
F = - 12 N
How do you interpret the minus sign??
um...12N toward the wall?
 
  • #6
The ball exerts an average 12 N force on and pointing toward the wall, so the wall must exert an average 12 N force on the ball in the direction, per Newton 3, pointing away from? or toward? the wall. The sign of the force corresponds to the signs you chose for the direction of the velocities.
 

1. What is impulse momentum theory?

Impulse momentum theory is a physics principle that describes the relationship between the force applied to an object, the time it is applied for, and the resulting change in momentum of the object.

2. How does impulse momentum theory apply to a ball striking a wall?

When a ball strikes a wall, the force of the impact and the time it takes for the impact to occur will determine the change in momentum of the ball. This change in momentum can be calculated using the impulse momentum equation, where impulse equals force multiplied by time.

3. What factors can affect the impulse and momentum in this scenario?

The force of the impact can be affected by the velocity of the ball, the mass of the ball, and the material of the wall. The time of the impact can be affected by the elasticity of the ball and the wall, as well as any external forces acting on the ball.

4. How is impulse momentum theory important in real-world applications?

Impulse momentum theory is used in various fields such as sports, engineering, and transportation. It helps in understanding the effects of forces on moving objects and can be used to design safer and more efficient structures and vehicles.

5. What are some limitations of impulse momentum theory?

Impulse momentum theory assumes that the force is constant throughout the entire impact, which may not always be the case. It also does not account for rotational motion or external factors such as air resistance. Additionally, it is only applicable to situations where the mass of the object remains constant.

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