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# Impulse and momentum dodgeball example

In this video, David shows how to solve for the impulse and force applied during a dodgeball collision using the impulse momentum relationship. Created by David SantoPietro.

## Want to join the conversation?

• Is this classified as an elastic collision?
• An elastic collision is one in which kinetic energy remains the same before and after the impact. The only way to determine whether this specific collision is elastic is the find the final and initial kinetic energy of the ball-face system. Since the formula for finding kinetic energy, KE = 1/2(m)(v^2), requires both mass and velocity, and neither the mass nor the initial velocity of the face are provided, it cannot be determined if this collision is elastic.
• What is basically an impulse? (Apart from its mathematical definition).
• Love the question dude!
anyways, this is just how I've understood impulse intuitively, and I'm just an 11th grader studying in India,

J is like the total increase/decrease in momentum over a given period of time (I know it's synonymous with J=change P, but read this through)
basically, what do we mean by a force of, say, 5N?
= The momentum is increasing by 5Ns per s, so the total increase in p over 2s = 5n*2 = 10N right?!
so over t seconds, it would be 5t Ns.
J = force acting over a period of time, what does this lead to, a change in momentum (over that period of time) hence the 2 have an equivalent formula.
so for the above case, we can say a force of 5 N exerted over a time period of 2s causes an impulse of 10Ns which is basically the total change in momentum
I know this is basic, but simplicity is (usually) the key,
I hope this appeals to your imagination as well,
if wrong, do let me know,
hope this helps :D
Onward!
• What about the air resistance?Doesn't air resistance affects the impulse?
• In most of the cases, while calculating such questions , Air resistance is usually neglected. If we consider air resistance, friction and such other non-conservative forces, we wont be able to apply conservation of energy / momentum in the problems as they make the system more complicated and prone to errors. If we are supposed to take air resistance under consideration, the value must be given in the question. Then we can calculate the force exerted by it and subtract it from the initial momentum. But still it will make the concept more complicated.
• Doesn't J stand for joules? What's the difference between J in impulse and J in joules?
• J for impulse and J for joules are used differently, like, J=Ft or 40J of energy. thus, you can always distinguish them and thus it doesn't matter. J for joules is a unit, J for impact is a vector quantity.
• So the final answer of the Second question (at the bottom) is +150N or -150N ?
• The answer would be -150 N because of the way the speaker formulated negatives and positives. The choice between which way is negative and which is positive is up to you, but just remember to be consistent. However, if the problem had asked for the magnitude impulse (hypothetically), the answer would always be 150 N, as magnitude means asking for the absolute value.
• What happens if the ball impacts at an angle and leaves at the same angle, do we need to consider that since Impulse is a vector?
• Hi I thought you multiply the net impulse force 3kg m/s by the time interval 0.02s but it is divided to give 150N, am I missing something?
• if you want to get an impulse given you know the net force and time interval, you can multiply them
: impulse = net_force * change_time

but here we know the net impulse (impulse is not a force, by the way) and time interval. thus we use the same formula above but with a bit of modification
: impulse = net_force * change_time
dividing both sides with change_time
impluse/change_time = net_force * change_time/change_time
which is same as
impluse/change_time = net_force
#change_time/change_time = 1

this is why we divide impulse (not a force, once again) by change_time to get a net force
(1 vote)
• At shouldn't it have been -5 instead of 5, as it comes back?
• No, in the problem the horizontal direction to the right was chosen as positive. In the equation the 10 m/s is negative and the 5 m/s is positive.

If you consider the initial direction of the balls movement, left, as positive then the 10 m/s is positive and 5 m/s would be negative.

Either is correct but you need to keep the directions labeled the same throughout the problem.