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3 years ago

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A 1200-kg SUV is moving alone a straight highway at 12.0 m/s. Another car, with mass 1800 kg and speed 20.0 m/s, has its center

of mass 40.0 m ahead of the center of mass of the SUV. Find (a) the position of the center of mass of the system consisting of the two cars; (b) the magnitude of the system’s total momentum, by using the given data; (c) the speed of the system’s center of mass; (d) the system’s total momentum, by using the speed of the center of mass. Compare your result with that of part (b).

1 answer:

andreev551 [17]3 years ago

8 0

Answer:

A) d = 24 m

B) 50400 kg.m/s

C) v₀ = 16.8 m/s

D) 50400 kg.m/s. It's equal to the momentum found in part B.

Explanation:

We are given;

Mass of station wagon;m1 = 1200 kg Velocity; V = 12 m/s

Mass of car; m2 = 1800 kg

Velocity of car; v2 = 20 m/s

a ) Let centre of mass of car and station wagon be at a distance d from wagon

Thus,

If we take moment of weight about it, we have;

1200 x d = 1800 x ( 40 - d )

Where, d is the position of the center of mass of the system consisting of the two cars

Thus,

1200d = 72000 - 1800d

1200d + 1800d = 72000

3000 d = 72,000

d = 72,000/3000

d = 24 m

b ) Total momentum= m1•v1 + m2•v2

= (1200 x 12) + (1800 x 20)

= 14400 + 36000

= 50400 kg.m/s

c ) Let speed of centre of mass be v₀

Thus,

v₀ = (m1•v1 + m2•v2)/(m1 + m2)

v₀ = 50400/(1200 + 1800)

v₀ = 50400/3000

v₀ = 16.8 m/s

d) System Total momentum = velocity of centre mass x total mass

Thus,

Total momentum = v₀(m1 + m2)

= 16.8(3000) = 50400 kg.m/s .

This value is equal to what was calculated in part b

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<u><em>La velocidad media es 5 </em></u> $\frac{km}{h}$ <u><em>, que equivale a 1.389 </em></u> $\frac{m}{s}$ <u><em></em></u>

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