2 Questions of Momentum and Impulse

In summary: For the problem about impulse I have J=F (delta t)=delta pF(t)=m(v2-v1)F(4.7*10^-3)=.145(52-40)J=.145*52+.145*40=1.74 (Kg*m)/sThe velocity components parallel and normal to the original velocity are:v1=v2=0v3=-(.145*52+.145*40)*cos(60.0)v4=-.145*52+.145*40*sin(60.0)
  • #1
wolfpack
3
0

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

ehild
 
  • #3
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
wolfpack said:
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
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
Find the velocity components parallel and normal to the original velocity. Apply sine and cosine of the given angle.

ehild
 

Related to 2 Questions of Momentum and Impulse

1. What is momentum?

Momentum is a physical quantity that measures the motion of an object. It is defined as the product of an object's mass and velocity. The direction of momentum is the same as the direction of the object's velocity.

2. How is momentum related to impulse?

Momentum is related to impulse through Newton's second law, which states that the change in an object's momentum is equal to the impulse applied to it. Impulse is the product of force and time, and can be calculated by integrating the force over time.

3. What is the difference between elastic and inelastic collisions?

In an elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy of the system before and after the collision are equal. In an inelastic collision, only momentum is conserved, and some kinetic energy is lost in the form of heat or sound.

4. How does the impulse-momentum theorem apply to real-life situations?

The impulse-momentum theorem can be applied to real-life situations such as car crashes. In the event of a car crash, the change in momentum of the car is equal to the impulse of the force exerted on it. This is why cars are designed with crumple zones to increase the time of impact and decrease the force exerted on the passengers.

5. How can impulse be used to increase the distance traveled by an object?

If the impulse applied to an object is constant, increasing the time of impact will also increase the change in momentum. This means that the object will have a smaller change in velocity and therefore travel a longer distance. This can be seen in sports such as long jump, where athletes use their momentum and apply a smaller force over a longer period of time to increase their distance traveled.

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