Elastic and Inelastic problems

[Kaneki]

Homework Statement

Im sorry to annoy anyone with this, i know to smart people these two are easy but I am dumb and i honestly don't understand this. If anyone can tell me and explain this concept then ill be very grateful. I have two days to learn and understand this and i don't know where else to go. So, if anyone can, please explain and direct me to where else i should try to understand this. Thank you so much

1 )With a mass of 114 kg Baby Bird is the smallest monoplane ever flown. Suppose the Baby Bird and pilot are coasting along the runway when the pilot jumps horizontally to the runway behind the plane. The pilot’s velocity upon leaving the plane is 5.32 m/s backward. After the pilot jumps from the plane, the plane coasts forward with a speed of 3.40 m/s. If the pilot’s mass equals 60.0 kg, what is the initial velocity of the plane and pilot before the pilot jumps?

2) The largest frog ever found was discovered in Cameroon in 1989. The
frog’s mass was nearly 3.6 kg. Suppose this frog is placed on a skateboard
with a mass of 3.0 kg. The frog jumps horizontally off the skateboard to
the right, and the skateboard rolls freely in the opposite direction with a
speed of 2.0 m/s relative to the ground. If the frog and skateboard are initially
at rest, what is the initial horizontal velocity of the frog?

Homework Equations

I think M1*V1 + M2*V2 = M1*V1 + M2*V2 and M1*V1 + M2*V2 = (M1*M2
)V2

The Attempt at a Solution

1). (60 + 114) = (114 * 5.32 + 60 * 3.40 )
174 606.48 + 204 = 810.48/174 = 6.5m/s

2.) 3.6 (2.0) + 3.0(0)
7.2 + 0
3.0 + 3.6 + 6.6, 7.2/6.6 = 1.0m/s

 

[Mickey Tee]

1) The plane had an initial velocity when the pilot was sitting in it.
As the pilot jumped backward with a certain momentum, an equal amount was imparted to the plane in the opposite direction, which is the direction of the plane.
This final momentum of the plane, divided by the mass of the plane equals 3.40m/s (which is greater than the initial velocity the plane had with the pilot sitting in it). Now work back to get the initial velocity of the plane from here.

2) The magnitude of the momentum of the frog (which jumps forward with v) equals the magnitude of momentum of the skateboards which has started moving in the opposite direction (you know both mass and velocity of skateboard)

 

[haruspex]

The pilot’s velocity upon leaving the plane is 5.32 m/s backward.

That's ambiguous, unfortunately. Is that relative to the ground or relative to the motion before jumping?

1). (60 + 114) = (114 * 5.32 + 60 * 3.40 )

Several problems there.
That equation only has constants. There should be an unknown somewhere.
Why are you multiplying the plane's mass by the pilot's speed, and vice versa?
The 5.32 and 3.40 are in opposite directions. Signs matter.

If the frog and skateboard are initially
at rest, what is the initial horizontal velocity of the frog?

Again, poor wording. Taken literally, the answer would be zero. It should say "with what velocity does the frog jump?"

2.) 3.6 (2.0) + 3.0(0)

That's not an equation, and you are again multiplying the mass of one thing by the speed of the other.