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WINSTONCH [101]
3 years ago
10

A 1000 kg red car traveling at 40 m/s rear ends a 3000 kg blue car traveling at 35 m/s. the cars bounce off each other and the b

lue car is accelerated to 37.5 m/s. what is the velocity of the red car after the crash?
Physics
1 answer:
klemol [59]3 years ago
8 0

Answer:

32.5 ms⁻¹

Explanation:

<em><u>Theory </u></em>

<u> The law of conservation of linear momentum </u>

The sum of linear momentum of a system under no external force remains constant.

In this scenario although different forces act on two vehicles by each other when we  consider the total system of two cars no external unbalance force act on them. So we can apply this law.

As law says,

The sum of linear momentum before collision = The sum of linear momentum after collision.

Momentum = mass × velocity

1000×40+3000×35 = 1000×v + 3000×37.5

where v = velocity of the red car after the cash

v = 32.5 ms⁻¹

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<h2>Answer:</h2>

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<h2>Explanation:</h2>

The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;

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y = (m₁y₁ + m₂y₂ + m₃y₃) / M       ----------------(ii)

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m₁ and x₁ = mass and position of first mass in the x direction.

m₂ and x₂ = mass and position of second mass in the x direction.

m₃ and x₃ = mass and position of third mass in the x direction.

y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.

From the question;

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Substitute these values into equations (i) and (ii) as follows;

x = ((6x0) + (4x3) + (2x0)) / 12

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x = 1 m  

y = (6x0) + (4x0) + (2x3)) / 12

y = 6 / 12

y = 0.5m

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