For this case we first think that the skateboard and the child are one body.
We have then:
1 = jug
2 = skateboard + boy
By conservation of the linear amount of movement:
M1V1i + M2V2i = M1V1f + M2V2f
Initial rest:
v1i = v2i = 0
0 = M1V1f + M2V2f
Substituting values
0 = (7.8) (3.2) + (M2) (- 0.65)
0 = 24.96 + M2 (-0.65)
-24.96 = (-0.65) M2
M2 = (-24.96) / (- 0.65) = 38.4 kg
Then, the child's mass is:
M2 = Mskateboard + Mb
Clearing:
Mb = M2-Mskateboard
Mb = 38.4 - 1.9
Mb = 36.5 Kg
answer:
the boy's mass is 36.5 Kg
From the equations of linear motion,
v² = u² + 2as where v is the final velocity, u is the initial velocity and a is the gravitational acceleration, and s is the displacement,
Thus, v² = u² -2gs, but v=0
hence, u² = 2gs
= 2×9.81×0.43
= 8.4366
u = √8.4366
=2.905 m/s
Hence the initial velocity is 2.905 m/s
Then using the equation v= u +gt .
Therefore, v = u -gt. (-g because the player is jumping against the gravity)
but, v = 0
Thus, u= gt
Hence, t = u/g
= 2.905/9.81
= 0.296 seconds
Looks like you need to review through the lesson and take notes as it tells you in the lesson what each of these are.
To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,
Here,
k = Coulomb's constant
= Charge at each object
r = Distance between them
As the distance is doubled so,
Therefore the factor is 1/4