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Impulse, elastic collisions and Inelastic...

by jrd007
Tags: collisions, elastic, impulse, inelastic
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jrd007
#1
Nov4-05, 06:27 PM
P: 159
Impulse
1) A gold ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5 x 10^-3 s. Find (a) the impulse imparted to the golf ball and (b) The average force exerted on the ball by the golf club.
Answer: (a) 2.0 kg(m/s) (b) 15.8 x 10^2 N

My approach....
(a) Impulse = velocity x change in mass = (0.045 kg)(45 m/s) = 2.0 kg(m/s) Got the first part.
(b) To get the average force would I not take impulse divided by time? 2.0/3.5 x 10^-3 = 571... but the answer is 15.8. x 10^2

----

Elastic Collisions
2) A 0.450 kg ice puck, moving east with a speed of 3.0 m/s has a head on collision with a 0.900 kg puck intially at rest. Assuming a perfectly elastic collision what will be the speed and direction of each object after the collision?
Answer: 1.00 m/s W & 2.00 m/s E

My thoughts were to just use the momentume formula...
(.5)(.450kg)(3 m/s) + (.5)(.900 kg)(0) = (.5)(.450kg)(v) + (.5)(.900 kg)(v)
But how do I determine it with two unknowns?

---

Inelastic Collisions
3) A 920 kg sports car collides into the rear end of a 2300 kg SUV stopped at a red light. The bumber's lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, knowing that the coeffiecent of kinetic friction b/t tires and road is 0.80 calculates the speed of the sports car at impact. What was that speed?
Answer: 23 m/s

Again I was going to use the same approach... but I had no speeds. Please someone help...
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Doc Al
#2
Nov4-05, 06:44 PM
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Quote Quote by jrd007
Impulse
1) A gold ball of mass 0.045 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 3.5 x 10^-3 s. Find (a) the impulse imparted to the golf ball and (b) The average force exerted on the ball by the golf club.
Answer: (a) 2.0 kg(m/s) (b) 15.8 x 10^2 N

My approach....
(a) Impulse = velocity x change in mass = (0.045 kg)(45 m/s) = 2.0 kg(m/s) Got the first part.
(b) To get the average force would I not take impulse divided by time? 2.0/3.5 x 10^-3 = 571... but the answer is 15.8. x 10^2
----
Your approach to (b) is correct; the given answer seems wrong.

Note: Impulse = change of momentum = mass x change in velocity.
Elastic Collisions
2) A 0.450 kg ice puck, moving east with a speed of 3.0 m/s has a head on collision with a 0.900 kg puck intially at rest. Assuming a perfectly elastic collision what will be the speed and direction of each object after the collision?

Answer: 1.00 m/s W & 2.00 m/s E
My thoughts were to just use the momentume formula...
(.5)(.450kg)(3 m/s) + (.5)(.900 kg)(0) = (.5)(.450kg)(v) + (.5)(.900 kg)(v)
But how do I determine it with two unknowns?
---
Don't forget to make use of conservation of energy since it's a perfectly elastic collision. (Also: Why do you have (.5) in front of each term?)

Inelastic Collisions
3) A 920 kg sports car collides into the rear end of a 2300 kg SUV stopped at a red light. The bumber's lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, knowing that the coeffiecent of kinetic friction b/t tires and road is 0.80 calculates the speed of the sports car at impact. What was that speed?
Answer: 23 m/s

Again I was going to use the same approach... but I had no speeds.
Since you have the stopping distance, you should be able to figure out the speed of the two cars just after the collision. (Figure out the friction force.)
jrd007
#3
Nov4-05, 07:28 PM
P: 159
By conservation of energy us mean PE and KE? I am still confused... elaborate please?

Doc Al
#4
Nov4-05, 07:37 PM
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Impulse, elastic collisions and Inelastic...

An elastic collision is one in which the total kinetic energy does not change during the collision. The objects bounce off of each other and no energy is "lost".

A perfectly inelastic collision, on the other hand, is one in which the objects stick together after colliding. The kinetic energy after the collision is less than before the collision; energy is "lost" (transformed into thermal energy and deformation).
jrd007
#5
Nov5-05, 09:45 AM
P: 159
Could I use p intial = p final?
m1v1 = (m1 + m2)v'
v' = [m1/(m1 +m2)](v1) = (.45/(.9 +.45))(3.0) = 1.00 m/s
An since I know that it is elastic and the sum of the two velocities will be 3 the other is 2.00 m/s.

Is the correct? And then I just assume that the direction will be west on the lighter object?

I am still having trouble with the last one. So how would I figure out my friction force again? which equation?
Doc Al
#6
Nov5-05, 10:46 AM
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Quote Quote by jrd007
Could I use p intial = p final?
Of course. That's conservation of momentum.
m1v1 = (m1 + m2)v'
v' = [m1/(m1 +m2)](v1) = (.45/(.9 +.45))(3.0) = 1.00 m/s
No. Here you assume that the final velocities of the masses are the same: that they stick together. But this is not true in an elastic collision. The conservation of momentum will give you this:
[tex]m_1 v_0 = m_1 v_1 + m_2 v_2[/tex]

An since I know that it is elastic and the sum of the two velocities will be 3 the other is 2.00 m/s.

Is the correct? And then I just assume that the direction will be west on the lighter object?
No, it's not correct at all. And there's no need to assume anything. (If it turns out that the lighter mass ends up traveling west, it's velocity will be negative.)

What you need to do is combine the conservation of momentum equation (given above) with conservation of energy:
[tex]1/2 m_1 v_0^2 = 1/2 m_1 v_1^2 + 1/2 m_2 v_2^2[/tex]
I am still having trouble with the last one. So how would I figure out my friction force again? which equation?
The force of kinetic friction is given by [itex]\mu N[/itex], where N is the normal force pushing the two surfaces together. (What's the normal force in this case?)
jrd007
#7
Nov5-05, 11:12 AM
P: 159
I have to combine the two equations... oh my. :( That is pretty difficult...

~~~

uN would be = (u)(weight of cars combined ?)
(0.80)(920 kg + 2300 kg) = 2576 N
Now that I have the force...
jrd007
#8
Nov5-05, 06:24 PM
P: 159
What next, Doc?
Doc Al
#9
Nov5-05, 07:22 PM
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Use the force to figure out the acceleration. (Newton's 2nd law.) Then use kinematics to find the initial speed of the two cars after the collision.
kp
#10
Nov5-05, 07:27 PM
P: 53
c) hmm...couldn't you say the K.E. of the collision 1/2(m1 + m2)vf^2 = the energy used by the force of friction u(m1+m2)gd. Then solve for vf to find the velocity of the cars after the collision?
jrd007
#11
Nov6-05, 08:05 AM
P: 159
So to figure out my acceleration, Doc, I take F = ma shifting into F/m = a.
2576 N/(920+2300 kg) = 2576/3220 = 0.8 m/s^2?

The I use kinematics... v2 = vo2 +2a(x-xo) 23 m/s
v^2 = 0 + 2(7.84 m/s^2)(2.8 m)
That defintely does not give me 23 m/s...
Doc Al
#12
Nov6-05, 08:19 AM
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Quote Quote by jrd007
That defintely does not give me 23 m/s...
No reason why it should. This is just the first step in the solution, not the final answer. Now that you've calculated the speed after the collision, use conservation of momentum to find the speed before the collision.
jrd007
#13
Nov6-05, 05:51 PM
P: 159
So it would be...

(920 kg)(v) + (2300 kg)(0) = (3220 kg)(6.63 m/s)
920 = 21349
= 23.21, wow. That was a LONG problem...
jrd007
#14
Nov7-05, 10:14 AM
P: 159
Now on to #2... This is waht I tried...

m1v1 = m1v1 + m2v2

v1 - v2 = - (v1 - v2)
3 = v1 + v2

1/2m1v1^2 = 1/2m1v1^2 + 1/2m2v2^2
Simplifies to: m1v1^2 = m1v1^2 + m2v2^2

3- v2 = v1
(3-v2)^2 = v1^2
9-6v2+v2^2 = v1^1
subsitute in...

4.05 = m1(9-6v2 + v2^2) + m2v2
4.05 = .45(9-6v^2 + v2^2) + .9 kg(v2)^2
4.05 = 4.05 - 2.7v2 + .45v2^2+.9v2^2
0 = 0 - 2.7v2 + 1.35v2^2
using the quad: a = 1.35, b = -2.7, c=0

The two answers I get are 2 and 0... the answers are really 2 and 1...
Doc Al
#15
Nov7-05, 11:44 AM
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Quote Quote by jrd007
Now on to #2... This is waht I tried...

m1v1 = m1v1 + m2v2
Careful. This should be:
[tex]m_1 v_0 = m_1 v_1 + m_2 v_2[/tex]

v1 - v2 = - (v1 - v2)
3 = v1 + v2
Not sure what this is supposed to be. Are you trying to make use of the fact that the relative velocity is reversed in an elastic collision? If so, then this should be:
[tex]v_A - v_B = v_B' - v_A'[/tex]
[tex]v_0 - 0 = v_2 - v_1[/tex]

Now you can combine this with the first equation and solve for v1 and v2. (Note: If you choose this method--highly recommended if you've covered it in class--then there's no need to apply conservation of energy separately: It's already included in this equation.)

1/2m1v1^2 = 1/2m1v1^2 + 1/2m2v2^2
Simplifies to: m1v1^2 = m1v1^2 + m2v2^2
Careful. This should be:
[tex]1/2 m_1 v_0^2 = 1/2 m_1 v_1^2 + 1/2 m_2 v_2^2[/tex]


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