In the game of billiards, all the balls have approximately the same mass, about 0.17 kg. In the figure, the cue ball strikes ano
ther ball such that it follows the path shown. The other ball has a speed of 1.5 m/s immediately after the collision. What is the speed (in m/s) of the cue ball after the collision?
1 answer:
Answer:
v = 2.6 m/s
Explanation:
The question is incomplete because the diagram is not given in the question.
The diagram of this question is attached below.
According to the law of conversation of momentum.
mv(initial) = mv(final)
We can see in the diagram that:
mv(initial) = 3 × 0.17
mv(final) = mv₁(final) + mv₂(final)
Consider only the horizontal components of velocities, to compare them to initial velocity.
mv(final) = mv₁(final) + mv₂(final)
0.17 × 3 = (0.17)(v)(cos 30) + (0.17)(1.5)(cos 60)
0.51 = 0.1472v + 0.1275
0.3825 = 0.1472v
v = 2.6 m/s
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Answer:
a) a = 3.06 10¹⁵ m / s , b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m - m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N
Answer:
0.003034 s
1.035 m
4.5 m
Explanation:
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