I need some help, I can't solve this physics problem.

Question

Answer 1

For a) Potential energy (E_P) is converted to kinetic energy E_K:

E_P1 = m₁gh = (2.00)(9.81)(5.00) = 98.1 joules

E_P1 = E_K1 so 1/2m₁v² = 98.1

1/2(2.00)v² = 98.1

v² = 98.1

v₁ = 9.9045 m/s = 9.90 m/s (to the right)

Same for block 2:

E_P2 = m₂gh = (4.00)(9.81)(5.00) = 196.2 joules

E_P2 = E_K2 so 1/2m₁v² = 196.2

1/2(4.00)v² = 196.2

v² = 98.1

v₂ = 9.90 m/s (to the left)

For b) set up a system of equation based on 1) conservation of momentum and 2) conservation of kinetic energy:

1) Conservation of Momentum (P_i = P_f):

Momentum just prior to the collision for block 1 (P_i):

P_1i + P_2i = m₁v_1i + m₂v_2i = (2.00)(9.9045) + (-4.00)(9.9045) = 19.809 - 39.618 = -19.809 kgm/s

The momentum immediately after the collision (P_f) is:

P_1f + P_2f = (2.00)v_1f + (4.00)v_2f = -19.809 kgm/s

2v_1f = -19.809 - 4v_2f

v_1f = -9.9045 - 2v_2f

2) Conservation of Energy (E_Ki = E_Kf)

Before the collision the total kinetic energy was 98.1 + 196.2 = 294.3 joules.

Right after the collision the total kinetic energy: 1/2m₁v_1f² + 1/2m₂v_2f² = 294.3 or

1/2(2.00)v_1f² + 1/2(4.00)v_2f² = 294.3

v_1f² + 2v_2f² = 294.3

We can substitute from the momentum equation v_1f = -9.9045 - 2v_2f so

(-9.9045 - 2v_2f)² + 2v_2f² = 294.3

98.1 + 39.618v_2f + 4v_2f² + 2v_2f² = 294.3

6v_2f² + 39.618v_2f - 196.2 = 0

Using the quadratic formula or other methods to solve the quadratic, we get v_2f = 3.3015 and -9.9045

Using v_1f = -9.9045 - 2v_2f, we get v_1f = -16.5075 and v_1f = 9.9045

So the two solutions are 1) v_1f = -16.5 m/s and v_2f = 3.30 m/s or 2) v_1f = 9.90 m/s and v_2f = -9.90 m/s. We can through out the second solution because the blocks can't pass each other by. That means block 1 is going 16.5 m/s left and block 2 is going 3.30 m/s right

For c), the E_P1 = E_K1 so:

m₁gh_1f = 1/2m₁v_1f²

(2.00)(9.81)h_1f = 1/2(2.00)(-16.5075)²

h_1f = 1/2(2.00)(-16.5075)²/(2•9.81) = 13.9 m

And E_P2 = E_K2 so:

m₂gh_2f = 1/2m₂v_2f²

(4.00)(9.81)h_1f = 1/2(4.00)(3.3015)²

h_2f = 0.556 m

So block 1 goes up to 13.9 m and block 2 goes to 0.556 m