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Figure 9.8,9 Impulse. Chapter 9 problems. b. What was the direction of travel before the explosion?. Chapter 9 problems. Chapter 9 problems. Chapter 9 problems. Chapter 9 problems. 3.0 m/s. 5.0 m/s.

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Chapter 9 problems
Chapter 9 problems

b. What was the direction of travel before the explosion?






3.0 m/s

5.0 m/s

  • Two balls of equal mass (m=0.15 kg) … Eventually the two balls collide, and after the collision the red ball is moving at 5.0 m/s and the blue ball is moving at 3.0 m/s.What was the change in momentum of the blue ball? What was the change in momentum of the red ball? Was momentum conserved in this collision? (most got this 21 right, 8 basically right but with a silly mistake (signs wrong, units wrong etc. ; 9 had various problems with units etc. 19 no answer)

  • It seems that there is no change in momentum of blue ball (also no change in momentum of red ball) becuase the total momentum before the collision is the same as the total momentum after the collision. The momentum was conserved in the collision. [BE CAREFUL]

  • The momentum was conserved, and the change in momentum for the blue ball was -.35N and the momentum change for the red ball was +.35N. [ONCE AGAIN WATCH UNITS AND NUMBERS]

  • blue ball = -0.3 kgm/s; red ball = +0.3 kgm/s ; yes, momentum was conserved [EXACTLY!]


  • Under what circumstances (if any) will linear momentum NOT be conserved in a physical process? (19 correct, 8 off slightly; 18 no answer)

  • if the collision is inelastic, linear momentum is mot conserved in the process. (Energy is not the issue here!)

  • there are no circumstances when linear momentum would not be conserved.

  • Linear momentum will not be conserved if a net external force acts on the system of momentum will not be conserved.

  • Momentum will be conserved to the extent that external forces are negligible compared to the forces between the interacting objects!


  • A red ball moves on a frictionless horizontal plane along the black line and collides with a stationary blue ball, the two balls then proceed along the red and blue lines respectively. At some time later, the two balls are at the positions marked by the circles with the black borders. Which ball has the greater mass?

  • (Equal:2 Red: 25 Blue: 4 Can’t tell: 4 No answer: 19)

  • They do not have equal masses or the red ball would've stopped and the blue ball would have then moved. The red ball has a greater mass because both balls moved after the collision. [Careful, it’s not a 1-D collision!!!]

  • from the picture I am assuming the red ball moves a shorter distance than the blue ball. I think that it is impossible tell because besides the mass, it depends on the velocity and angle of impact. But i'm really not sure, maybe it can be deduced.



x is always between the balls and moves in a straight line. Since it moves in a straight line, the center of mass must be closer to the red ball given the final positions so the red ball is more massive.

Now let’s analyse this

Problem more completely

Vby = 5 Vry

pbx = pry

Vrx ~ 6 Vry

(Can’t really compare initial and final speeds directly from the figure since time information not given).

y

  • The key is to focus on the momentum perpendicular to the incident momentum!

  • Clearly (I think) the y component of the blue ball’s velocity is greater than the red’s. Since the y component of the total momentum must be zero after the collision, the red ball MUST be more massive.


Figure 9 8 9 impulse1
Figure 9.8,9 Impulse is always between the balls and moves in a straight line. Since it moves in a straight line, the center of mass must be closer to the red ball given the final positions so the red ball is more massive.


Calm inquiries
Calm inquiries is always between the balls and moves in a straight line. Since it moves in a straight line, the center of mass must be closer to the red ball given the final positions so the red ball is more massive.

  • Energy Conservation/ PE (5 requests)

  • Collisions (3)

  • Angles (2 people)


Chapter 9 problems5
Chapter 9 problems is always between the balls and moves in a straight line. Since it moves in a straight line, the center of mass must be closer to the red ball given the final positions so the red ball is more massive.


Chapter 8 problems
Chapter 8 problems is always between the balls and moves in a straight line. Since it moves in a straight line, the center of mass must be closer to the red ball given the final positions so the red ball is more massive.


Basic ideas for sheet
Basic ideas (for sheet) is always between the balls and moves in a straight line. Since it moves in a straight line, the center of mass must be closer to the red ball given the final positions so the red ball is more massive.

  • Work = F.d W = DK Power = dW/dt (=F.v)

  • Kinetic Energy: K = ½ mv2

  • P.E.: Ug = mgh Uel = ½ k(x-xo)2

  • F = -dU/dx (e.g. Spring: F = -k(x-xo))

  • Mech. Eng. conserved if only conservative forces do work. Dissipative forces convert some Mech. Eng. to thermal energy.

  • Center of Mass R = S(miri)/Smi

  • Impulse: J=DP = FavDt = integral of F(t)dt

  • P= mvFnet= dP/dt => momentum is conserved if outside forces can be neglected compared to internal forces.

  • Defns: Elastic, Perfectly inelastic, etc. for collisions, Conservative forces, dissipative forces etc. BEWARE of applying 1-D ideas to multiple dimensions (or elastic equations to partially inelastic problems etc.).


Office hours next week
Office Hours Next Week is always between the balls and moves in a straight line. Since it moves in a straight line, the center of mass must be closer to the red ball given the final positions so the red ball is more massive.

  • DVB

    • Friday (today): 1:30 to 3:00

    • Monday: 9:00 to 10:00 as usual,

      and 1:00 to 2:30

    • Tuesday: 3:00 to 5:00

  • Aditi:

    • Monday: 2:30 to 3:30


y is always between the balls and moves in a straight line. Since it moves in a straight line, the center of mass must be closer to the red ball given the final positions so the red ball is more massive.

V3=?

V1=3.50m/s

q=45o

x

x

q=35o

V2=?

Vo=5.00m/s

  • . A 4.00 kg circular container is sliding across a frictionless surface along the x-axis in the positive direction at a speed of 5.00 m/s as shown below on the left. As it crosses the origin, a small internal “explosion” splits the container into three pieces (or mass 2.00 kg, 1.00kg and 1.00kg each). The two lighter fragments travel at 45.0o and 35.0o to the negative x-axis with the first of these is travelling at 3.50 m/s (see figure on the right). The heaviest fragment continues travelling along the positive x-axis after the explosion

    • What is the vector momentum of the container prior to the explosion?

    • What is the combined vector momentum of the three fragments after the explosion?

    • What is the speed of the second of the two lighter fragments after the collision?

    • What is the speed of the heaviest fragment after the explosion?

    • What is that minimum value for the energy of the “explosion” in order for the above description to be physically possible?


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