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marissa [1.9K]
3 years ago
13

A 1200 kg car traveling north at 10 m/s is rear-ended by a 2000 kg truck traveling at 30 m/s. What is the total momentum before

and after the collision?
Physics
1 answer:
tino4ka555 [31]3 years ago
4 0

Answer:

The total momentum before and after collision is 72000 kg-m/s.

Explanation:

Given that,

Mass of car = 1200 kg

Velocity of car = 10 m/s

Mass of truck = 2000 kg

Velocity of truck = 30 m/s

Using conservation of momentum

The total momentum before the collision is equal to the total momentum after collision.

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})V

Where, m_{1}=mass of car

v_{1} =velocity of car

m_{1}=mass of truck

v_{1} =velocity of truck

Put the value into the formula

1200\times10+2000\times30=(1200+2000)V

V=\dfrac{1200\times10+2000\times30}{(1200+2000)}

V = 22.5\ m/s

Now, The total momentum before collision is

P=m_{1}v_{1}+m_{2}v_{2}

P=1200\times10+2000\times30

P=72000\ kg-m/s

The total momentum after collision is

P=(m_{1}+m_{2})v_{2}

P=(1200+2000)\times22.5

P= 72000 kg-m/s

Hence, The total momentum before and after collision is 72000 kg-m/s.

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A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° no
Rzqust [24]

Answer:

a)  R = 2.5 mi   b)  To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

     Cos 20 = Aₓ / A

     sin 20 = A_{y} / A

    Aₓ = A cos 160 = 2 cos 160

    A_{y}  = A sin160 = 2 sin160

    Aₓ = -1,879 mi

    A_{y}  = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

    cos 2600 = Bₓ / B

    sin 260 = B_{y} / B

    Bₓ = B cos 260

     B_{y}  = B sin 260

    Bₓ = 4 cos 260

     B_{y}  = 4 sin 260

     Bₓ = -0.6946mi

     B_{y}  = - 3,939 mi

Third vector C = 3 mi to 15 north east

     cos 15 = Cₓ / C

     sin15 = C_{y} / C

     Cₓ = C cos 15

     C_{y} = C sin15

     Cₓ = 3 cos 15

    C_{y} = 3 sin 15

     Cₓ = 2,898 mi

    C_{y} = 0.7765 mi

Now we can find the final position of the person

    X = Aₓ + Bₓ + Cₓ

    X = -1.879 -0.6949 + 2.898

    X = 0.3241 mi

    Y = A_{y} +  B_{y} + C_{y}

    Y = 0.684 - 3.939 +0.7765

    Y = -2.4785 mi

a) We use Pythagoras' theorem

     R = √ (x2 + y2)

     R = √ (0.3241 2 + (-2.4785) 2)

     R = 2.4996 mi

     R = 2.5 mi

b) let's use trigonometry

     Tan θ = y / x

     Tanθ  = -2.4785 / 0.3241

     θ = tan⁻¹ (-7,647)

     θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

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In this problem, we have

F = 15 N

d = 2.0 m

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Substituting into the equation, we find

\tau = (15)(2.0) sin 90^{\circ}=30 N \cdot m

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