Answer:
C. In the first collision has twice the momentum as when it stays still ( second colllions)
Explanation:
To see which statement is correct, it is best to solve the problem, the momentum is equal to the variation of the moment
I = Δp = m vf - m v₀
I = m (vf -v₀)
Case 1. In car bounces, the initial speed is 0.3 m / s, say that this direction is positive, when the magnitude of the speed bounces it remains constant, but its direction is reversed (vf = -0.3 m / s)
I₁ = m (-0.3 - 0.3)
I₁ = -0.6 m
Case 2. The expensive one that still after the crash so its speed is zero (vf = 0)
I₂ = m (0 - 0.3)
I₂ = -0.3 m
Let's calculate the relationship between the two impulses
I₁ / I₂ = -0.6m / -0.3m
I₂ / I₂ = 2
When it bounces it has twice the momentum as when it stays still
Now let's analyze the answers:
A. False The momentum changes
B. False. The momentum is less in the second collision
C. True. The momentum is double in this collision
D. False. Can be calculated, because the mass is the same throughout the exercise and is eliminated in the equations
E. False. When they say bounces it implies the same speed with the opposite direction
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