Solve Impulse Problem: Average Force on Bat

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In summary, Daniel hit a fastball at 120 mph that had a mass of 150 g. The ball had a contact time of .001 seconds. The average force on the bat was 9120 kgm/s.
  • #1
Azytzeen
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A baseball player hits a 82 mph fastball, sending it back at 120 mph. The ball has a mass of 150 g. The contact time is 0.001 s. What is the average force on the bat?

Okay, so I converted the miles to kilometres, and then used the equation F_avg*(t_2-t_1) = m_1*v_2 - m_2*-v_2, but I can't get the right answer. I even eliminated the negative sign and tried that, but it is still wrong. I can't see what else to add, so please help.
 
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  • #2
Assuming you meant [itex]F_{ave} \Delta t = \Delta (mv) = m (v_f - v_i)[/itex], and you did your unit conversions properly, that should work. Note that if [itex]v_i = + 82\ \mbox{mph}[/itex], then [itex]v_f = - 120\ \mbox{mph}[/itex]. (Signs matter, since momentum is a vector.)
 
  • #3
[tex] \bar{\vec{\mbox{F}}}=:\frac{\vec{\mbox{p}}_{f}-\vec{\mbox{p}}_{i}}{\Delta t} [/tex]

U know the momentum both initially and finally and u know the time of impact (in which the momentum is being transfered).

Daniel.
 
  • #4
82*1.6=131.2
120*1.6=192
.15*192-.15*131.2=9.12
9.12/.001=9120

That's the answer I got the first time, but the computer says that it is wrong. Did I do everything right?
 
  • #5
Check the units.U need to convert everything to SI-mKgs.

Daniel.
 
  • #6
82*1.6=131.2 km
120*1.6=192 km
.15kg*192km-.15kg*131.2km=9.12kg*km
9.12kg*km/.001sec=9120
If I convert it into kgm/s it becomes 9.12*10^6N. Hmm, still wrong... Argh!
 
  • #7
Azytzeen said:
82*1.6=131.2
120*1.6=192
It looks like you're converting miles to km; what you should be doing is converting miles/hour to meters/second.
.15*192-.15*131.2=9.12
Reread my comments about signs. Realize that the ball reverses direction. For example: if it comes towards the bat at 10 mph, then leaves the bat at 15 mph, the change in velocity would be: 15 - (-10) = 25 (not 15 - 10 = 5).
 
  • #8
Initial momentum (negative by a choise of axis) [tex] -0.15 \ \mbox{Kg} \cdot \frac{(82\cdot 1.6) \cdot 1000 \ \mbox{m}}{3600 \ \mbox{s}} [/tex]

Final momentum (positive) [tex] +0.15\ \mbox{Kg} \cdot \frac{(120\cdot 1.6)\cdot 1000 \ \mbox{m}}{3600 \ \mbox{s}} [/tex]

Compute the 2 #-s and then subtract the negative one from the positive one.The result should be divided by the time interval.

Daniel.
 
  • #9
Oh... damnnit! Thanks guys, I will try that out now.
 

Related to Solve Impulse Problem: Average Force on Bat

1. What is an impulse problem?

An impulse problem involves the calculation of the change in momentum of an object over a specific period of time. In the context of a bat, it refers to the force applied to the bat by a ball during a collision.

2. How do you calculate the average force on a bat during a collision?

The average force on a bat can be calculated by dividing the change in momentum of the bat by the time it takes for the collision to occur. This can be represented mathematically as F = Δp/Δt, where F is the average force, Δp is the change in momentum, and Δt is the time interval.

3. What factors affect the average force on a bat during a collision?

The average force on a bat can be affected by several factors, including the mass and velocity of the ball, the elasticity of the bat, and the angle of impact between the bat and the ball. Other factors such as air resistance and friction may also play a role.

4. How can the average force on a bat be increased?

In order to increase the average force on a bat during a collision, the mass and velocity of the ball can be increased, or the angle of impact can be changed to be more perpendicular to the surface of the bat. Additionally, using a bat with a higher elasticity can also result in a higher average force.

5. Why is it important to solve impulse problems in sports?

Solving impulse problems in sports, such as the average force on a bat during a collision, can help athletes and coaches understand the mechanics and physics behind their performance. It can also aid in identifying areas for improvement and optimizing techniques for better results.

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