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March 19th, 2017, 12:30 AM   #1
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Help with conservation of momentum question

Q1) Tiny Tim of mass 72kg is playing in the super bowl against the Arizona Cardinals. Tiny Tim is running at 4ms-1 when he collides with Alan Branch, with mass 165kg, running in the opposite direction with a speed of 5ms-1.
Momentarily after the collision Alan Branch’s velocity was 0ms-1.

ii. Calculate Tiny Tim’s speed after the collision




iii. If the collision took place over a time of 0.3 seconds, what was the force experienced by Tiny Tim during the collision?



iv. Hence or otherwise, what was the force experienced by Alan Branch during the collision?
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March 19th, 2017, 05:17 AM   #2
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Show what you have attempted ...
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March 19th, 2017, 12:51 PM   #3
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I got the final velocity of Tiny Tim, which was -7.46ms^-1

I am stuck on question iii and iv as I'm not quite sure how to go about it.

Last edited by skipjack; March 19th, 2017 at 10:45 PM.
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March 19th, 2017, 02:12 PM   #4
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Quote:
Originally Posted by Posher View Post
I got the final velocity of Tiny Tim, which was -7.46ms^-1
correct

Quote:
iii. If the collision took place over a time of 0.3 seconds, what was the force experienced by Tiny Tim during the collision?
impulse equation ...

$F \cdot \Delta t = m \cdot \Delta v$


Quote:
iv. Hence or otherwise, what was the force experienced by Alan Branch during the collision?
Apply Newton's 3rd law.
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Last edited by skipjack; March 19th, 2017 at 10:45 PM.
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March 19th, 2017, 09:08 PM   #5
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Also, for iv, wouldn't the force received from Alan be 0 since the final velocity is 0?

Edit: for iii, I got -1790.4N

Last edited by skipjack; March 19th, 2017 at 10:41 PM.
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March 20th, 2017, 07:23 AM   #6
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(ii) Tiny Tim's speed is $|v| = |-7.46 \, m/s| = 7.46 \, m/s$

(iii) $F \Delta t = m \Delta v \implies F = \dfrac{m(v_f-v_0)}{\Delta t} = \dfrac{72(-7.46 - 4)}{0.3} = -2750 \, N$

(iv) same equation for impulse on the larger player ...

$F \Delta t = M \Delta v \implies F = \dfrac{M(v_f-v_0)}{\Delta t} = 2750 \, N$


Quote:
wouldn't the force received from Alan be 0 since the final velocity is 0?
No. This is a common misconception related to another misconception ... "if acceleration = 0 then velocity = 0", or the converse of that statement.

Force is the agent of change. That force acting on an object for a defined period of time yields an impulse, which by another name is change in momentum.

The force on the larger player is equal and opposite in direction to the force acting on the smaller player. Heard that before?

Just because velocity = 0 does not imply that force = 0.
An example ... think about a ball thrown straight up under the influence of gravity. At the top of its trajectory, the velocity is 0, but the force of gravity acts on it the entire time it moves as a projectile in the gravitational field.

Another example ... a car drives at high speed and hits a wall bringing it to rest in minimal time. Did the wall exert a force on the car causing it to stop? Did the car exert an equal and oppositely directed force on the car?
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Last edited by skeeter; March 20th, 2017 at 07:30 AM.
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March 20th, 2017, 01:35 PM   #7
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(iii) $F \Delta t = m \Delta v \implies F = \dfrac{m(v_f-v_0)}{\Delta t} = \dfrac{72(-7.46 - 4)}{0.3} = -2750 \, N$



shouldn't the -4 be positive since Tim is traveling in opposite directions
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March 20th, 2017, 03:29 PM   #8
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Tim ...

$v_0 = +4 \, m/s$, $v_f = -7.46 \, m/s$

$\Delta v = v_f - v_0 = -7.46 - (+4) = -11.46 \, m/s$

Branch ...

$v_0 = -5 \, m/s$, $v_f = 0$

$\Delta v = v_f - v_0 = 0 - (-5) = 5 \, m/s$


kapish?
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March 20th, 2017, 07:58 PM   #9
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I already it figured that out like 10 minutes after posting the reply. I'm sorry I wasted your time in posting another reply :/
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March 21st, 2017, 03:09 AM   #10
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Quote:
Originally Posted by Posher View Post
I already it figured that out like 10 minutes after posting the reply. I'm sorry I wasted your time in posting another reply :/
It was not a waste. I enjoyed reading this thread and I think others enjoyed reading this thread too.

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