The formulae and concepts used here are:
p=mv,
change in p in time Dt
is the impulse FDt.
The vector momentum is always conserved in isolated systems.
1. The total momentum before the dive = 0, which = the total after the dive. The mvraft must be equal and opposite to the mvdiver (Newton's third law).
2. The only way to conserve both momentum and energy (in an elastic head-on collision) is if the second ball carries off all the momentum and the first ball remains at rest.
3. Depending on the sophistication of the reader/writer of this question, the answer is either (a) or (c). In all real collisions (except between some elementary particles) some energy always goes into heat and vibrations.
5. The change in momentum is 1.5 N*s, so the force is 1.5/0.025 N.
7. One can use dimensional analysis for this one. Notice that 4.5 g/s times 80 m/s has the units grams*meters/sec2, which is mass*accel. So the answer is 4.5*80/400 = 0.9 m/s2.
The "physics class" solution is:
Every second, the change in momentum (impulse,
=FDt) is 0.0045kg * 80m/s, so
the force applied =
Dp/Dt =
0.36 N, and the resulting acceleration is F/m = 0.9 m/s*s.
8. The change in air-momentum every second is m*40, so the force is m*40/1 (the /1 sec keeps the proper units) which must be equal to mcopterg.
9. We are given BOTH the angle AND the speed of
of ball#1 after the collision.
In class I did not read the problem carefully enough!
Given the after-speed, we do not need conservation of energy.
Ball#1 has positive 2.17 m/s y-motion.
The balls must have equal and opposite y-components
of velocity, (since there was no y-momentum to begin with),
so Ball#2 has negative 2.17 m/s y-motion.
After the collision, the sum of the two x-momenta must equal the initial
x-momentum. Since the two masses are equal, the sum of the x-velocities
must equal the initial x-velocity of 5 m/s.
Ball#1 has x-motion positive 3.77 m/s, so Ball#2
has x-motion 1.23 m/s.
Now we know both components of the 2nd ball's post-collision velocity.
Pythagoras and tangents will provide the answer.