1. It is important to remember that in this situation, the projectile has an intial horizontal motion of 30 m/s, and an initial vertical motion of 0 m/s. In flight, there are no horizontal forces on the projectile (if we neglect air friction). This means that there are no accelerations in the horizontal direction. Thus, the horizontal motion of the projectile is constant; this means that it moves horizontally at 30 m/s as long as it is in flight. This them implies that the horizontal distance of the projectile from the cliff is easily determined according to: dist = speed x time, since there is no acceleration in the horizontal direction. Thus, the projectile moves forward 30 m/s for each sec it is in the air; at t = 1 sec, the projectile is 30 m from the edge of the cliff, at t = 2 sec it is 60 m away, at t = 3 s it is 90 m from the cliff, at t = 4 s it is 120 m away, and at t= 5s it is 150 meters away.

The vertical motion is different in that there is an acceleration in the vertical direction, namely the acceleration due to gravity. So, the vertical speed of the projectile increases by 10 m/s for each sec of flight. Hence, the vert speed at t =1 sec is 10 m/s, at t=2 s is 20 m/s. In general, the vert speed is simply gt, where g is the accel due to gravity (10 m/s/s) and t is time in secs.

The vertical distance is determined from d = 1/2 gt². You have a couple of ways you can report the vertical distance; you can report this as how far below the edge of the cliff the object has dropped, or you can report this as height above the ground. In either event, you need to calculate that after 1 sec, the object has dropped:

The total speed at each time is determined from the Pythagorean Theorem; that is:
v_total² = v_hor² + v_vert²

DATA TABLE FOR QUESTION 1

t(sec)	V hor (m/s)	V vert (m/s)	hor dist (m)	vert dist (m)	total speed (m/s)
0	30	0	0	0	30
1	30	10	30	5	31.6
2	30	20	60	20	36.1
3	30	30	90	45	42.4
4	30	40	120	80	50
5	30	50	150	125	58.3

3.To walk on the log, you must exert a backward force on the log; this causes the log to move backward. You could also argue from momentum conservation: all forces are internal to the person/log system, so if the person moves forward, the log must move backward to conserve total momentum.

4. Well, technically, the Earth and satellite are in mutual orbit. The point is that the forces are equal. This means that the force on the satellite equals the force on the Earth. However, the Earth has a much, much greater mass than the satellite, so the same force acting on the two objects causes a great acceleration on the satellite, and an infinitesimally small one on the much more massive Earth.

5. This is really the same problem as above. The forces on the bug and bus are equal, however, the masses are different, so the bug accelerates much more than the bus.

6. a. Momentum = mass x velocity = 8 kg x 2 m/s = 16 kgm/s
b. Force = change in momentum/change in time. We know from above that the momentum is 16kgm/s. After the impact, the momentum goes to zero, so the change in momentum is -16 kgm/s (remember, change in momenmtum = momentum_final - momentum_initial). Dividing this by the time in which it takes to bring the momentum to zero (0.5s) yields an average force of 32N. (this is also the answer to part c.)

7. Impulse = F x delta t = F x delta (mv). So, impulse = 10N x 2.5 s = 25 Ns.
b) Since Impulse = change in momentum, 25 kgm/s is the change in momentum.
c) If we are told that the cart was originally at rest, then we know the change in momentum is equal to the final momentum. Hence we know that mv_final = 25kg m/s. If the mass is 2 kg, then v_final = 12.5 m/s.

8. a) This is an example of a conservation of momentum problem. They system consists of the two blobs of putty. The momentum before collision is the sum of the momenta of the two blobs: total momentum = 2kg x 3 m/s + 2 kg x 0 m/s = 6 kgm/s.
The momentum after collision is the momentum of the combined mass: 4kg x v_final.
Since momentum is conserved, i.e., does not change, the momentum before collision must equal the momentum after collision and we obtain:

9. a) True, this is simply Newton's third Law.
b) True, part a) tells us the forces are the same, and the collision lasts the same length of time as seen by the car or by the bug.
c) No, the masses are very different so the speeds are different.
d) True, if impulses are the same (see part b)), then the changes in momentum are the same.

10. a) Remember that impulse = F x delta t = change in momentum. We know that the momentum before the collision is 20,000 kg m/s, and is zero after collision (the car comes to rest). Hence, we know the change in momentum is 20,000 kg m/s, hence this is the value of impulse.
b) We know that F x delta t = 20,000 kg m/s, but without knowing delta t, we cannot compute F.

11. This is a conservation of momentum problem. Before collision, (i.e., eating), the total momentum of the system is:

If the smaller fish were swimming toward the larger one at 4 m/s, then the initial momentum of the system is: