To solve this problem, we must remember about the law of
conservation of momentum. The initial momentum mist be equal to the final
momentum, that is:
m1 v1 + m2 v2 = (m1 + m2) v’
where v’ is the speed of impact
Since we are not given the masses of each car m1 and m2,
so let us assume that they are equal, such that:
m1 = m2 = m
Which makes the equation:
m v1 + m v2 = (2 m) v’
Cancelling m and substituting the v values:
50 + 48 = 2 v’
2 v’ = 98
v ‘ = 49 km/h
<span>The speed of impact is 49 km/h.</span>
Answer:
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Answer:
6.77 m/s
Explanation:
First, in the x direction:
Given:
Δx = 3.17 m
v₀ = v cos 30.8° = 0.859 v
a = 0 m/s²
Δx = v₀ t + ½ at²
(3.17 m) = (0.859 v) t + ½ (0 m/s²) t²
3.17 = 0.859 v t
3.69 = v t
Next, in the y direction:
Given:
Δy = 0.432 m
v₀ = v sin 30.8° = 0.512 v
a = -9.81 m/s²
Δy = v₀ t + ½ at²
(0.432 m) = (0.512 v) t + ½ (-9.81 m/s²) t²
0.432 = 0.512 v t − 4.905 t²
Two equations, two variables. Solve for t in the first equation and substitute into the second equation:
t = 3.69 / v
0.432 = 0.512 v (3.69 / v) − 4.905 (3.69 / v)²
0.432 = 1.89 − 66.8 / v²
66.8 / v² = 1.458
v² = 45.8
v = 6.77
Answer:
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