Two masses collide inelastically.

Question

Two masses collide inelastically. Before the collision mass 1 has a velocity of 14.7 m/s at 43.3^o and mass 2 has a velocity of 13.7 m/s at 21.6^o. Following the collision mass 1 has a velocity of 11.1 m/s at 61.5^o. If mass 1 is 10.7 kg and mass 2 is 9.9 kg, what are the speed and direction of mass 2 following the collision?

Umm, I have no idea what is going on here. I have tried making a diagram and solving for the speed and direction but am lost. Collisions are just a bunch of scrambled gobbledygook in my head, if someone can explain this step by step with a diagram I would really approach it.

Answer 1

Momentum is conserved in the x-direction and momentum is conserved in the y-direction.

Let:

m₁ = mass of mass 1

m₂ = mass of mass 2

v_1-b = velocity of mass 1 before the collision

v_1-bx = velocity of mass 1 before the collision in the x-direction

v_1-by = velocity of mass 1 before the collision in the y-direction

v_2-b = velocity of mass 2 before the collision

v_2-bx = velocity of mass 2 before the collision in the x-direction

v_2-by = velocity of mass 2 before the collision in the y-direction

v_1-a = velocity of mass 1 after the collision

v_1-ax = velocity of mass 1 after the collision in the x-direction

v_1-ay = velocity of mass 1 after the collision in the y-direction

v_2-a = velocity of mass 2 after the collision

v_2-ax = velocity of mass 2 after the collision in the x-direction

v_2-ay = velocity of mass 2 after the collision in the y-direction

P_bx = Momentum before the collision in the x-direction

P_1bx = Momentum of mass 1 before the collision in the x-direction

P_2bx = Momentum of mass 1 before the collision in the x-direction

P_by = Momentum before the collision in the y-direction

P_1by = Momentum of mass 1 before the collision in the y-direction

P_2by = Momentum of mass 1 before the collision in the y-direction

P_ax = Momentum after the collision in the x-direction

P_1ax = Momentum of mass 1 after the collision in the x-direction

P_2ax = Momentum of mass 1 after the collision in the x-direction

P_ay = Momentum after the collision in the y-direction

P_1ay = Momentum of mass 1 after the collision in the y-direction

P_2ay = Momentum of mass 1 after the collision in the y-direction

Calculations for Momentum before the collision

v_1-b = 14.7 (at 43.3°) therefore:

v_1-bx = 14.7cos(43.3°) = 10.7

v_1-by = 14.7sin(43.3°) = 10.1

v_2-b = 13.7 (at 21.6°) therefore:

v_2-bx = 13.7cos(21.6°) = 12.7

v_2-by = 13.7sin(21.6°) = 5.04

Momentum in the x-direction before collision

P_bx = P_1bx + P_2bx = (m₁)•(v_1-bx) + (m₂)•(v_2-bx) = (10.7)(10.7) + (9.9)(12.7) = 114.5 + 126.1 = 240.6

Momentum in the y-direction before collision

P_by = P_1by + P_2by = (m₁)•(v_1-by) + (m₂)•(v_2-by) = (10.7)(10.1) + (9.9)(5.04) = 107.9 + 49.9 = 157.8

Calculations for Momentum after the collision

v_1-a = 11.1 (at 61.5°) therefore:

v_1-ax = 11.1cos(61.5°) = 5.30

v_1-ay = 11.1sin(61.5°) = 9.76

v_2-a = ? (at θ°) therefore:

v_2-ax = v_2-acos(θ)

v_2-ay = v_2-asin(θ)

Momentum in the x-direction after collision

P_ax = P_1ax + P_2ax = (m₁)•(v_1-ax) + (m₂)•(v_2-ax) = (10.7)(5.30) + (9.9)(v_2-acos(θ)) = 56.7 + 9.9v_2-acos(θ)

Momentum in the y-direction after collision

P_ay = P_1ay + P_2ay = (m₁)•(v_1-ay) + (m₂)•(v_2-ay) = (10.7)(9.76) + (9.9)(v_2-asin(θ)) = 104.4 + 9.9v_2-asin(θ)

Conservation of Momentum in the x-direction

P_bx = P_ax

240.6 = 56.7 + 9.9v_2-acos(θ)

183.9 = 9.9v_2-acos(θ)

18.6 = v_2-acos(θ)

v_2-a = 18.6/cos(θ)

Conservation of Momentum in the y-direction

P_by = P_ay

157.8 = 104.4 + 9.9v_2-asin(θ)

53.4 = 9.9v_2-asin(θ)

5.40 = v_2-asin(θ)

Solving

since v_2-a = 18.6/cos(θ), plug "18.6/cos(θ)" in for v_2-a in the 2nd equation 5.40 = v_2-asin(θ):

5.40 = (18.6/cos(θ)•sin(θ)

5.40/18.6 = sin(θ)/cos(θ)

Using the identity tan(θ) = sin(θ)/cos(θ):

5.40/18.6 = tan(θ)

0.290 = tan(θ)

θ = tan^-1(0.290)

θ = 16.2°

Plug 16.2° into v_2-a = 18.6/cos(θ):

v_2-a = 18.6/cos(16.2°)

v_2-a = 19.3 m/s