Week 2 Lecture 5
Solutions at the bottom.

1)  In-video quiz at timepoint 03:28.

The small steel balls in the experiment Dr. Plesser just showed will ______ .

a) continue moving because the momentum of the group is conserved. 
b) eventually stop because air friction is interacting with the group.

2)  In-video quiz at timepoint 09:29.

Conservation of Energy: If the bowling ball drops a height of 0.8 meters from Dr. Plesser's chin to its lowest point, how fast is it traveling at its lowest point, assuming there is no friction? Give the speed in meters per second, rounded to two significant digits.

[Open-ended]

3)  In-video quiz at timepoint 09:29.

Given the previous speed of the bowling ball, what would be the momentum for a 5 kg (11 lb) bowling ball at the lowest point of its path? Give your answer in kg⋅m/s, rounded to one significant digit.

[Open-ended]

4)  In-video quiz at timepoint 09:29.

Conservation of momentum: Say that Dr. Plesser persuaded his guileless TA Justin to stand at the lowest point of the bowling ball's path. If the 90 kg TA absorbs all of the bowling ball's momentum, how fast will he be knocked backwards? Give your answer in meters per second rounded to two significant digits.

[Open-ended]

Solutions:

1) Correct answer is d) 2 m/s2 south

The change in velocity is 6 meters per second. The change in time is 3 seconds. Therefore the magnitude of the acceleration is 6 / 3 = 2 meters per second squared (m/s2 ). The direction of the acceleration is south because the north-moving car is slowing down. If the car had sped up to 16 meters per second, the magnitude of the acceleration would have been the same, but the direction would have been north, not south.\

2) Correct answer is (3.9,4.1)

3) Correct answer is (19.5,20.5)

Momentum is mass times velocity. 5kg⋅4m/s=20kg⋅m/s

4) Correct answer is (0.21,0.23)

Initial momentum is 20 kg⋅m/s, from the previous problem. Final momentum is 90kg⋅vTA=20kg⋅m/s. Solving for the velocity, we get vTA=0.2222m/s, which rounds to 0.22 meters per second.