# Kinetics Problem: Finding average normal force

1. Nov 28, 2009

### woody89

I'd just like to say thank-you in advance for any and all answers. It is greatly appreciated.

1. The problem statement, all variables and given/known data

The 2kg ball is thrown at the suspened 20kg block with a velocity of 4m/s. If the time of impact between the ball and the block is 0.005s, determine the average normal force exerted on the block during this time. Take e=0.8.

2. Relevant equations

mava+mbvb=mava'+mbvb'

e=vb'-va'/va-vb

mv1+Ft=mv2

3. The attempt at a solution

First I designated a to denote the ball and b to denote the block

Using mava+mbvb=mava'+mbvb'

I know that because the block is at rest and therefore vb=0m/s
Subbing in the values given in the problem statement I was able to create an equation for vb' and va'

(2)(4)+(0)=(2)va' + (20)vb'
once simplified I had 4=va'+10vb' --->eq1

Then using e=vb'-va' /va-vb and the values given in the problem I was able to get another equation for vb' and va'

0.8= vb'-va' /2
1.6=vb' -va' ---->eq2

eq1+eq2 gave me vb'=0.509m/s which I then used to find va'=2.12m/s

Then using mv1+Ft=mv2 and va'=2.12m/s as v2 I calculated for F
F=(2)(2.12)-(4)(2)/0.005 = -756.36 N
This value of F seemed very large to me given that the ball is travelling as a speed of only 4m/s.

Because my assignments are done online I know, as I tried pluging in this value, that that is not the correct answer. My problem is that I have no idea where it is that I went wrong. Any direction on this would be greatly appreciated.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 28, 2009

### ehild

"0.8= vb'-va' /2" is wrong. (vb'-va')/4 = 0.8.

ehild

3. Nov 28, 2009

### kuruman

Recheck your work. The momentum before the collision is
Pbefore= 2*4 = 8.0 kg m/s
If your calculation is correct, the momentum after is
Pafter=2*2.12 + 20*0.509 = 14.4 kg m/s
and your numbers imply that momentum is not conserved.

Without seeing your exact calculation, it is hard to figure out where you went wrong, but I see two sticky points with your expression for the coefficient of restitution.
1. Why do you say 0.8= vb'-va' /2 ?
The denominator should be 4 - 0 = 4, not 2.

2. If the ball bounces back after the collision, then it must be a negative number and the relative velocity in the numerator becomes the sum of the speeds, not the difference.

Once you have the correct velocities, your method for figuring out the force is correct.

4. Nov 28, 2009

### woody89

Okay so I tried with the new formula over 4. It was a stupid mistake I made thanks for catching it :). But when I recalculated with that change I got a force of -581.8N. I plugged that into the website and it was wrong. Because I had used up all my attempts I found out the answer was 2.62 kN. Unfortunately I'm still lost on how they got that.

5. Nov 29, 2009

### ehild

The correct eq2 is

vb'-va'=3.2

eq1: va'+10vb'=4

Add them together: 11 vb'=7.2--->vb'=0.6545 m/s.

Calculate the change of momentum of the block, (20*0.6545) divide by the time during the change happened, the result is 2618 N.

ehild