EXAMPLE 1: Two air carts have total masses m1=320 grams and m2=195 grams. Assume that there are absolutely, positively, undoubtedly, unequivocally no frictional or other external, horizontally directed forces acting on the system of the two air carts. Initially, cart #1 is moving to the right and cart #2 is moving to the left. Cart #1 has a speed of 0.324 m/s, and Cart #2 has a speed of 0.766 m/s. Then the two carts collide and stick together. Cart 1 Photogates Cart 2 a. Using your knowledge of physics, determine the expected momentum of the system AFTER the collision. Make sure you designate which direction is positive. Remember units! (-0.0457 kg m/s, with positive being to the right) b. What is the total kinetic energy before the collision? Again, remember your units! (0.0740 J) c. Consider the KE after the collision. In this ideal experiment, do you expect it to go down, go up, or remain the same? Why? (It goes down, because anytime objects collide and stick together, KE always goes down. Method 2: Use momentum conservation to find that the velocity afterward is -0.0887 m/s, then use the KE formula to find KE of 0.00203 J. This is indeed far less KE than there was before the collision.)

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Collisions in 1-D: Linear Momentum and KE
Be able to describe the types of collisions (elastic and inelastic) and whether or
not momentum and/or kinetic energy is conserved for each of these.
Given the masses of 2 carts on an ideal frictionless airtrack, and the velocity
either before or after a collision, calculate the system momentum and kinetic energy, or
the individual momentum (with sign) and KE of a cart.
Given the system's momentum or kinetic energy before the collision, predict that
the system's momentum and KE should be the same after an elastic collision. However, if
the carts stick together, the momentum should be the same though the KE must go down.
Suppose you were given complete data for the carts before the collision, and
partial data for the carts after the collision (or vice versa). Be able to use the conservation
of momentum principle to solve for unknowns such as the velocity (speed and direction)
of either cart, or the mass of either cart.
EXAMPLE 1: Two air carts have total masses m1=320 grams and m2=195 grams.
Assume that there are absolutely, positively, undoubtedly, unequivocally no frictional or
other external, horizontally directed forces acting on the system of the two air carts.
Initially, cart #1 is moving to the right and cart #2 is moving to the left. Cart #1 has a
speed of 0.324 m/s, and Cart #2 has a speed of 0.766 m/s. Then the two carts collide and
stick together.
Cart 1
Photogates
Cart 2
a. Using your knowledge of physics, determine the expected momentum of the system
AFTER the collision. Make sure you designate which direction is positive. Remember
units! (-0.0457 kg m/s, with positive being to the right)
b. What is the total kinetic energy before the collision? Again, remember your units!
(0.0740 J)
c. Consider the KE after the collision. In this ideal experiment, do you expect it to go
down, go up, or remain the same? Why? (It goes down, because anytime objects collide
and stick together, KE always goes down. Method 2: Use momentum conservation to
find that the velocity afterward is -0.0887 m/s, then use the KE formula to find KE of
0.00203 J. This is indeed far less KE than there was before the collision.)
EXAMPLE 2: A 210-gram cart moves right at a speed of 5.00 m/s and collides with an
unknown mass car moving left at a speed of 4.00 m/s. They stick together and move
toward the left at a speed of 0.576 m/s. What is the unknown mass? (342 grams) What is
the momentum before the collision? (-0.318 kg m/s) What is the % change in the
system's KE? (-98.3%)
Transcribed Image Text:Collisions in 1-D: Linear Momentum and KE Be able to describe the types of collisions (elastic and inelastic) and whether or not momentum and/or kinetic energy is conserved for each of these. Given the masses of 2 carts on an ideal frictionless airtrack, and the velocity either before or after a collision, calculate the system momentum and kinetic energy, or the individual momentum (with sign) and KE of a cart. Given the system's momentum or kinetic energy before the collision, predict that the system's momentum and KE should be the same after an elastic collision. However, if the carts stick together, the momentum should be the same though the KE must go down. Suppose you were given complete data for the carts before the collision, and partial data for the carts after the collision (or vice versa). Be able to use the conservation of momentum principle to solve for unknowns such as the velocity (speed and direction) of either cart, or the mass of either cart. EXAMPLE 1: Two air carts have total masses m1=320 grams and m2=195 grams. Assume that there are absolutely, positively, undoubtedly, unequivocally no frictional or other external, horizontally directed forces acting on the system of the two air carts. Initially, cart #1 is moving to the right and cart #2 is moving to the left. Cart #1 has a speed of 0.324 m/s, and Cart #2 has a speed of 0.766 m/s. Then the two carts collide and stick together. Cart 1 Photogates Cart 2 a. Using your knowledge of physics, determine the expected momentum of the system AFTER the collision. Make sure you designate which direction is positive. Remember units! (-0.0457 kg m/s, with positive being to the right) b. What is the total kinetic energy before the collision? Again, remember your units! (0.0740 J) c. Consider the KE after the collision. In this ideal experiment, do you expect it to go down, go up, or remain the same? Why? (It goes down, because anytime objects collide and stick together, KE always goes down. Method 2: Use momentum conservation to find that the velocity afterward is -0.0887 m/s, then use the KE formula to find KE of 0.00203 J. This is indeed far less KE than there was before the collision.) EXAMPLE 2: A 210-gram cart moves right at a speed of 5.00 m/s and collides with an unknown mass car moving left at a speed of 4.00 m/s. They stick together and move toward the left at a speed of 0.576 m/s. What is the unknown mass? (342 grams) What is the momentum before the collision? (-0.318 kg m/s) What is the % change in the system's KE? (-98.3%)
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