# Homework Help: Impulse problem.

1. Apr 10, 2005

### Azytzeen

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.

2. Apr 10, 2005

### Staff: Mentor

Assuming you meant $F_{ave} \Delta t = \Delta (mv) = m (v_f - v_i)$, and you did your unit conversions properly, that should work. Note that if $v_i = + 82\ \mbox{mph}$, then $v_f = - 120\ \mbox{mph}$. (Signs matter, since momentum is a vector.)

3. Apr 10, 2005

### dextercioby

$$\bar{\vec{\mbox{F}}}=:\frac{\vec{\mbox{p}}_{f}-\vec{\mbox{p}}_{i}}{\Delta t}$$

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

Daniel.

4. Apr 10, 2005

### Azytzeen

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. Apr 10, 2005

### dextercioby

Check the units.U need to convert everything to SI-mKgs.

Daniel.

6. Apr 10, 2005

### Azytzeen

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. Apr 10, 2005

### Staff: Mentor

It looks like you're converting miles to km; what you should be doing is converting miles/hour to meters/second.
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. Apr 10, 2005

### dextercioby

Initial momentum (negative by a choise of axis) $$-0.15 \ \mbox{Kg} \cdot \frac{(82\cdot 1.6) \cdot 1000 \ \mbox{m}}{3600 \ \mbox{s}}$$

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

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. Apr 10, 2005

### Azytzeen

Oh... damnnit! Thanks guys, I will try that out now.