Answer:
The carriage speed B was 67.4 mph
Step-by-step explanation:
This is an exercise for the moment, that as a vector we must look for the solution of each axis (x, y). We define a system formed by the two cars, for this system the forces during the crash are internal, so the moment is preserved.
The data they give is the car more A m = 1000kg and its speed is v1₁₀ = 30 mph i^ and the mass of the car B M = 1500 kg
Let's write the moment for each axis
X axis
p₀ₓ =
![p_(fx)](https://img.qammunity.org/2020/formulas/physics/college/66eb2tn437rgud2w3n8lyxkuyrngrw6rbn.png)
m v₁ₓ + 0 = (m + M) vₓ
Y axis
poy =
![p_(fy)](https://img.qammunity.org/2020/formulas/physics/college/m8acp3jrziwz54vwgmtx8opaf7vdq9karv.png)
0 + M
= (m + M)
Let's look for the components of the final velocity with trigonometry
sin 66 =
/ v
cos 66 = vₓ / v
= v sin 66
vₓ = v cos 66
We substitute and write the system of equations
m v₁ₓ = (m + M) v cos 66
M
= (m + M) v sin66
From the first equation
v = m / (m + M) v₁ₓ / cos 66
v = 1000 / (1000 + 1500) 30 / cos 66
v = 29.5 mph
From the second equation
= (m + M)/m v sin 66
= (1000 + 1500) /1000 29.5 sin 66
= 67.4 mph
The carriage speed B was 67.4 mph