Help on AP Physics C question?

Good Question. First we define the coordinates. Let j be the unit vector along the positive y axis.

a) Relative to the ground, the coin moves at 9.8m/s^2 towards the ground. In unit vector notation we write -9.8 j m/s^2. This is because the compartment (and the customer who is tied to the compartment) falls faster than g, thus "leaving the coin behind" to fall at its natural acceleration of g.

b) Relative to the customer, who is travelling at -1.24g j, the coin is travelling at an acceleration of

- g j - (-1.24g) j = -g j + 1.24g j = 0.24g j. (Because a vector B relative to a vector A, is B - A)

c) From part b), relative to the compartment, the coin is accelerating upwards at a rate of 0.24g. (This is because the customer and compartment speeds are equal). The coin must travel a distance of 2.2m relative to the compartment. We wish to determine the time. We thus use the equation:

s = ut + 0.5*at^2

where s = distance, u = initial velocity (in this case 0), a = acceleration, t = time.

Therefore we have,

2.2 = 0 + 0.5*0.24*g*t^2

Therefore,

2.2 = 0.12*9.8*t^2

Therefore,

2.2 = 1.176t^2

Therefore,

1.87 = t^2

Therefore,

t = 1.367 seconds

d) Force = mass*acceleration

= 0.567*-g j

=- 0.567*9.8 j

= -5.56 j Newtons

e) Apparent force = mass*apparent acceleration relative to customer

= 0.567*0.24*g j

= 0.567*0.24*9.8 j

= 1.33 j Newtons

Since one newton is [(kg*m)/s^2] you should convert the 0.567 g coin to 5.67x10^-4 kg