Understanding Momentum and Impulse

[Sonny18n]

Homework Statement

1) A hockey player makes a slap shot, exerting a force of 30.0N on the hockey pock for 0.16 seconds. What impulse is given to the puck?

2) The hockey puck shot in exercise 1 has a mass of 0.115 kh and was at rest before the shot. With what speed does it head toward the goal.

Homework Equations

p=mv

(F)triangle(t)= Triangle(p)

The Attempt at a Solution

1. If I'm using impulse formula correctly than multiplying 30N and 0.16 sec is 4.8

2) I have a harder time understanding this, I'm not even sure if it's asking me to use the formula for momentum[/B]

 

[DrClaude]

1. If I'm using impulse formula correctly than multiplying 30N and 0.16 sec is 4.8

Units?

2) I have a harder time understanding this, I'm not even sure if it's asking me to use the formula for momentum

What is the relation between momentum and speed?

 

[dean barry]

The solution can be found starting with with Newtons classic:
f = m * a
Think about the definition for acceleration (a).

 

[haruspex]

The solution can be found starting with with Newtons classic:
f = m * a
Think about the definition for acceleration (a).

Yes, but given the first part of the question I would think the intent is to use momentum.

 

[Sonny18n]

Ok so how do I find the acceleration

 

[haruspex]

Ok so how do I find the acceleration

You don't need to. See DrClaude's post.

 

[Sonny18n]

You don't need to. See DrClaude's post.

That's all the question says, the "at rest" part is kind of throwing me off

 

[haruspex]

That's all the question says, the "at rest" part is kind of throwing me off

Why? Suppose it moves off at speed v. How much is the increase in speed? What is the increase in momentum?

 

[dean barry]

Force * time gives the value for impulse.
The result equals the change in momentum, during the impulse.
The original momentum is 0

 

[haruspex]

Force * time gives the value for impulse.
The result equals the change in momentum, during the impulse.

Sonny18n already did that bit in part 1. DrClaude and I are trying to get Sonny18n to use that result together with the puck mass to solve part 2 directly rather than going back to the force given. I (we?) feel this is the intent of the question, and clearly something Sonny18n needs to learn how to do. The force * time approach is not always available, because it only works when the force is constant. Yes, it's constant here, but in other momentum questions it won't be.

 

[Sonny18n]

Why? Suppose it moves off at speed v. How much is the increase in speed? What is the increase in momentum?

At rest means 0m/s right? So it won't affect the equation in a major way. But it's asking me to relate the answer from question 1 but it's not clear to me how impulse I can find the speed it takes.

 

[haruspex]

At rest means 0m/s right? So it won't affect the equation in a major way. But it's asking me to relate the answer from question 1 but it's not clear to me how impulse I can find the speed it takes.

Impulse is momentum. How do you normally assess the momentum of an object?

 

[Sonny18n]

At rest means 0m/s right? So it won't affect the equation in a major way. But it's asking me to relate the answer from question 1 but it's not clear to me how impulse I can find the speed it takes.

Impulse is momentum. How do you normally assess the momentum of an object?

Mass times velocity?

 

[haruspex]

Mass times velocity?

Yes.

 

[Sonny18n]

Yes.

So the velocity is 0m/s then or..?

 

[haruspex]

So the velocity is 0m/s then or..?

If the momentum is mass times velocity, what do you think a change in momentum for an object equals?

 

[Sonny18n]

If the momentum is mass times velocity, what do you think a change in momentum for an object equals?

But I don't have momentum. Only mass and no velocity

 

[haruspex]

But I don't have momentum. Only mass and no velocity

You have change in momentum and mass, and you want change in velocity.

 

[yuganes warman]

the first question is asking about impusle which the formula is ft=mv-mu. So in order to get the impulse just multiply the force given and the time. You may ask where does ft=mv-mu came from, it is derived from the F=ma. Acceleration equals to v-u/t. So when yiu substitute v-u/t into the acceleration , you will get F=mv-mu/t. The t is taken to the left side of the equation and it becomes Ft=mv-mu. mv-mu is the impulse and so ft is also impulse

 

[MacLaddy]

Keep in mind that momentum is conserved, meaning that it can not change unless there is an outside influence acting on it- an impulse.

 

[Chestermiller]

Your two equations tell the whole story. If you combine them, you get:
$$FΔt=Δ(mv)$$
What is F and Δt in your problem?
What is mv to begin with?
You are trying to find the final value of v from this equation. You do know how to express Δ(mv) algebraically in terms of the initial and final velocities of the puck, correct?

Chet

 

[dean barry]

You could go by another route, use Newtons classic relationship: f = m * a
( transpose for acceleration a )
Then find the velocity after time t
Calculate the momentum.
( m * v )