water balloon toss.
INTRODUCTION - IMPULSE - fill the blanks.
Imagine 2 eggs dropped from the same height. One is to fall on the ground and the second one on a flexible net.
Before reaching their destination, they both are in free-fall, so they both have the same ___________ V.
(we neglect frictional forces).
The net (le'ts call this egg lucky) or the floor (let's call this egg humpty) is going to stop the egg. So the final speed = _______ for both eggs.
The difference is in the time of impact. It will takes longer to stop the egg falling on the ________
than the one falling on the ______. Call t1 the time to stop egg lucky and t2 the time to stop
egg humpty we have t1 ___ t2 (is it = or < or > ? ). Both eggs have the same mass m .
According to Newtons second law : m Δv = F t (m Δv is the change in momentum, F t is called the impulse)
Δv is the change in velocity, here Δv = ____ (final = 0, initial = V), m is the mass, F is the force the egg exerts on the floor or
the net when it hits it and t the time of the impact.
For egg lucky , using the formula, we have F egg on net t1= _________ (use V, t1, m)
For egg humpty we have F egg on floor t2= ___________ (use V, t2, m)
So They both undergo the same change in ___________________. (same change in state of motion)/
So we have F egg on net t1 = F egg on floor t2 (the impulses are the same)
But because t1 >> t2 then __________ << _____________
REmember Newton's third law ? Flucky on net = - Fnet on lucky and or course Fhumpty on floor = - Ffloor on humpty
(action reaction) so Fnet on lucky ____ Ffloor on humpty (is it << ? or >> ? or = ? )
This means that as you increase the time of impact, you decrease the _________ acting on the object about to stop.
And this for the same impulse, for the same change in momentum, for the same change in state of motion.
Newton's second law says: m Δv = F t
m Δv is the change in ___________
and F t is the _______. For F t to stay constant (that is m Δv to stay constant), you can either decrease F if you
increase the _____ or increase ______ if you decrease the time of impact t.
In this lab we will demonstrate how the time can affect the force and therefore the damage applied to an object
that is decelerating to a stop. You will throw water balloon and try not to break them by increasing the time of impact.
PROCEDURE. The winner team is the one who can keep their water balloons as long as possible.
READ FIRST:
Each lab group should fill a balloon with equal amount of water and then exit the school to a playing field.
A member of each group should stand 2 meters away from his or her partner and gently toss the balloon so the partner will be able to catch it without breaking.
After each successful catch, take another step back and toss it again. As the distance between partners increases it is necessary to throw the balloon with greater speed and momentum. it is necessary to increase the time over which the halting force is applied if the momentum is to be reduced to zero without increasing the force applied on the balloon. The winners often minimize the force on the balloon by extending their arms forward and, as the balloon lands, moving them backward so tithe balloon decelerates slowly.
OBSERVATIONS:
Who was the winner ? what was their technique ?
ANALYSIS
1) What did you learn ?
2) Bungee divers have to use flexible bungee cords and they bounce back 10 or 15 times before coming to rest. Explain why it would be dangerous to use ropes instead ? (think newton's second law)
3) Which technique is most effective in catching the water-balloon ? Explain in terms of momentum and impulse ?
4) A swimmer who jumps from a high diving board can enter the water and sustain no injury. The same person jumping from the same height and landing on concrete may not live to tell about it. Explain in terms of impulse and momentum.
5) Suppose you are traveling in a bus at highway speed on a nice Summer day and a unlucky bug is suddenly smashed against
your windshield.
A) Compared to the force that acts on the bug, how much force F acts on the bus ? (think Newton's third law)
B) The time of impact t is the same for the bus and the bug. Compared to the impulse on the bug, this means the impulse on the bus is :
more ? the same ? less ? hint: impulse = force x time = F t
C) remember impulse = change in momentum or F t = m (∆ V) or F t = change in momentum
So ,although the momentum of the bus is very large compared to the momentum of the bug, the change in the momentum of
the bus compared to the change of momentum of the bug is : more ? the same ? less ?
D) so change in momentum bug = change in momentum bus
or m ∆
So for the change in momentum to be the same, given the relative mass of the colliders, the change in speed of the bug has to be huge,
and the change in speed of the bug really tiny. The effect of this change in momentum is really different.
Which one will survive the collision?
Which undergoes the greater acceleration ?
Which therefore, suffers the greater damage?