Lab Quiz 5

Question 1: (1 point)

The sketch above shows a small glider of 162.3 grams colliding with a large stationary glider of 243.3 grams on the left. On the right the two gliders are shown in motion after the collision. The velocities of the gliders after the collision are 0.06248 m/s for the small glider and 0.300 m/s for the large glider. What is the momentum of the incident glider? Indicate with a positive (negative) sign whether the small glider moves right (left) after the collision. (NOTE! Contrary to the first part of your lab, the collision is NOT necessarily elastic, i.e. the equations in Ch7 sheet 29 do not necessarily hold, so don't use them. Use momentum conservation only (see Ch7 sheet 10), which holds for all collisions, whether elastic or inelastic. The direction of motion of the small glider after the collision is important for the choice of the sign of its velocity, when you apply momentum conservation. Use the positive sign for the incident direction.)

(How to Solve):

Question 2: (1 point)

The sketch above shows a large glider of 266.1 grams colliding with a small stationary glider of 165.9 grams on the left. On the right the two gliders are shown in motion after the collision, stuck together. The velocity of the two stuck-together gliders after the collision is 0.134 m/s. How much kinetic energy is lost in this perfectly inelastic collision (see Ch7 sheet 14'), i.e. what is the difference of the total kinetic energies before and after the collision? (Hint: first calculate the velocity of the incident glider from momentum conservation (on sheet 14' consider only the x-components). Then calculate the difference asked for.) Indicate with a negative (positive) sign whether the two stuck-together gliders always move to the right (can sometimes move to the left).

(How to Solve):


Question 3: (1 point)

The sketch above shows a large glider of 264.1 grams colliding with a small stationary glider of 163.8 grams on the left. On the right the two gliders are shown in motion after the collision, stuck together. The velocity of the two stuck-together gliders after the collision is 0.119 m/s. The velocity of the stuck-together gliders after the collision is measured with a photogate that records the time, during which a small flag mounted on top of the gliders blocks the light beam of the photo gate while the gliders move through the gate. You neglect the error in the time measurement and you have estimated that the 5 cm width of the flag has a 2 mm error. What is the absolute error of the total kinetic energy after the collision if the error of the two masses is 1 gram each? (Hint: get the relative error of the velocity from the relative error of the flag, using the fact that the time error is neglected. Then use extression (6), ( and (7) in "Error and Uncertainty".) Indicate with a positive (negative) sign whether the contribution of the mass errors to the overall error is large (very small).

(How to Solve):

Anyone know how to do the lab quiz questions? please HELP?

I am really stuck on these if anyone could help that would be great.

1) The sketch above shows a small glider of 164.2grams colliding with a large stationary glider of 241.7grams on the left. On the right the two gliders are shown in motion after the collision. The velocities of the gliders after the collision are 0.05709m/s for the small glider and 0.300 m/s for the large glider. What is the momentum of the incident glider ? Indicate with a positive (negative) sign whether the small glider moves right (left) after the collision. (NOTE! Contrary to the first part of your lab, the collision is NOT necessarily elastic, i.e. the equations in Ch7 sheet 29 do not necessarily hold, so don't use them. Use momentum conservation only (see Ch7 sheet 10), which holds for all collisions, whether elastic or inelastic. The direction of motion of the small glider after the collision is important for the choice of the sign of its velocity, when you apply momentum conservation. Use the positive sign for the incident direction.)

(.1642 * .05709)- (.2417 * .3)
My answer was -.06314

Does anyone know how to get number 2 or 3?

Any ideas on questions 2 and 3?

First use the formula:
v'=(m_1*v_1)/(m_1+m_2) and solve for v_1 which gives:
v_1=(v'*(m_1+m_2)/m_1 this gives you the initial velocity for the large glider.

Remember the KE =.5*m*v^2 ?
Plug in the velocity that you just solved for and the large mass that is given to find the initial KE.

Then use the final velocity (given) and m_1+m_2 for the mass in the same equation for KE as above.

subtract both values of KE to find the difference of the two KE's and that is your answer

Don't know the logic but I did get the pathway to the answer:

relative error of the flag = relative error of the velocity so..
0.002/0.05= 0.04
equation #8 gives 2*0.04=0.08
use KE = .5*m*v^2 as above for the KE AFTER the collision
0.08 * your answer for KE
I am off to MapleTA and then to bed. I hope I was of some assistance.

Ok so I did that for question 3 but it marked it as wrong.. does anyone have another way?

Ok nvm, the way to do it is right, I didn't realize the problem had different numbers.

This is confusing, can someone write out the step by step solution for Question 3? Thank you

Here's how to do #3:

ΔKE=sqrt((ΔM+m/M+m)^2 + (2 * Δv/v)^2) * KE

Δv/v= Δd/d = 0.05/0.002

ΔM+m= sqrt((0.001)^2 + (0.001)^2), get M+m from the masses given in the problem.

I need help I tried the equation but it doesn't work

are numbers 2 and 3 positive or negative?

I tried numbers 2 and 3 twice and I keep getting them wrong as well...

is this pos. or neg?