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2 Questions of Momentum and Impulse

  • Thread starter wolfpack
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  • #1
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Homework Statement


Problem 1: A .145 kg baseball pitched at 40 m/s is hot on a horizontal line back to the pitcher at 52 m/s. If the contact time of the ball with the bat is 4.7e-3, calculate the magnitude of the average force acting on the ball due to the bat.

Problem 2:A billiard ball is moving at 3.1 m/s when it hits a stationary ball of the same mass. After the collision, the second ball moves at 2.4 m/s at an angle of 60.0 deg to the original line of motion, Find the magnitude and direction of the velocity of the first ball after the collision.

Homework Equations


Problem1) Impulse= change in momentum=m(v1-v2)=F((delta t))
Problem2)p=m*v


The Attempt at a Solution


My attempts at a solution are shown in the jpegs, with enumerated variables and equations I've tried. As for the method behind the madness, For problem one I am going off of the face that since impulse is the change is momentum, I can obtain a value from the quantifiable values for the momenta and then set the impulse equal to F delta t isolating the force in question.
For problem 2 I am going off of the conservation of momentum in a perfectly elastic collision. I know the enter of mass are uniform and masses are the same, and then I tried solving for the components of the momenta.
These questions are from the Andes OLI Learn by Doing series.
Thank you for your assistance
 

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Answers and Replies

  • #2
ehild
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I can not follow your calculations, what have you got for the change of momentum in the first problem?

ehild
 
  • #3
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For the problem about impulse I have J=F (delta t)=delta p
F(t)=m(v2-v1)
F(4.7*10^-3)=.145(52-40)
Thus F=(.145(52-40))/(4.7*10^-3)=370.213 N
Apparently that is incorrect, but I am unsure of exactly what other equations to use with the quantities I have.

The change in momentum I got was (.145*52-.145*40)=1.74 (Kg*m)/s
 
  • #4
ehild
Homework Helper
15,477
1,854
For the problem about impulse I have J=F (delta t)=delta p
F(t)=m(v2-v1)
F(4.7*10^-3)=.145(52-40)
You forgot that the initial and final velocities have opposite directions. If v1 is positive, v2 is negative. Think it over.

ehild
 
  • #5
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Ah I see what I did now, so with the revised computation the correct solution is 2838.298 N which I checked and it works. Thank you

I still have that momentum equation left for my OLI, I'm just rusty on collisions in two dimension in general. I know the resultant angles have some connection to the component angles of the momenta, but I'm really just not sure where to proceed.
 
  • #6
ehild
Homework Helper
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1,854
Find the velocity components parallel and normal to the original velocity. Apply sine and cosine of the given angle.

ehild
 

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