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A 750kg truck traveling at 5m/s east collides with a 1500kg car moving at 20m/s in a direction of 30 degree south of west .after collision the two vehicles remain tangled together . with that speed and in what direction does the wreckage begin to move ?

Answer 1

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JGottem1189

01/20/2020

Physics

College

+10 pts

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A 7500 kg truck traveling at 5m/s east collides with a 1500kg car moving at 20m/s in a direction 30 degrees south of west. After collision, the two vehicles remain tangled together. With what speed and in what direction does the wreckage begin to move?

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CarliReifsteck

Answer:

The speed and direction of the wreckage is 2.09 m/s and 52.74°.

Explanation:

Given that,

Mass of truck = 7500 kg

Speed of truck = 5 m/s

Mass of car = 1500 kg

Speed of car = 20 m/s

Angle = 30°

We need to calculate the velocity after collision along east

Using conservation of momentum

m_{t}v_{t}+m_{c}v_{c}\cos\theta=Mv_{east}

Put the value into the formula

7500\times5-1500\times20\cos30=(7500+1500)v_{east}

v_{east}=\dfrac{7500\times5-1500\times20\cos30}{7500+1500}

v_{east}=1.27\ m/s

We need to calculate the velocity after collision along north

Using conservation of momentum

m_{t}v_{t}+m_{c}v_{c}\cos\theta=Mv_{north}

Put the value into the formula

7500\times0-1500\times20\sin30=(7500+1500)v_{north}

v_{north}=\dfrac{7500\times0-1500\times20\sin30}{7500+1500}

v_{north}=-1.67\ m/s

The resultant velocity is

v=\sqrt{(v_{east})^2+(v_{north})^2}

Put the value into the formula

v=\sqrt{(1.27)^2+(-1.67)^2}

v=2.09\ m/s

We need to calculate the direction

Using formula of direction

\theta=\tan^{-1}(\dfrac{v_{north}}{v_{east}})

Put the value into the formula

\theta=\tan^{-1}(\dfrac{|v_{north}|}{|v_{east}|})

\theta=\tan^{-1}(\dfrac{1.67}{1.27})

\theta=52.74^{\circ}

Hence, The speed and direction of the wreckage is 2.09 m/s and 52.74°.

Answer 2

Answer:

The wreckage begins to move with a speed of 3.15m/s and in a direction of 66.04°.

Explanation:

Given:

Mass of truck ($m_t$) = 7500kg

Speed of truck ($v_t$) = 5m/s

Mass of car ($m_c$) = 1500kg

Speed of car ($v_c$) = 20m/s

α = 30°

To find:

Speed (v) and direction (θ) of the wreckage.

Step 1:

In the east direction:

The momentum before the collision is equal to the momentum after the collision, i.e., momentum is conserved.

Therefore, by the conservation of momentum, we get

$m_tv_t-m_cv_ccos\alpha = Mv_e$

Substituting the given values,

$(7500*5) - (1500*20)cos(30) = (7500+1500)v_e$

$v_e= \frac{37500-25980}{9000}$

$v_e=\frac{11520}{9000}$

$v_e = 1.28m/s$

Step 2:

In the north direction:

The momentum before the collision is equal to the momentum after the collision, i.e., momentum is conserved.

Therefore, by the conservation of momentum, we get

$m_tv_t-m_cv_ccos\alpha = Mv_n$

Substituting the given values,

$(7500*0) - (1500*20)cos(30) = (7500+1500)v_n$

$v_n= \frac{0-25980}{9000}$

$v_e=\frac{-25980}{9000}$

$v_e = -2.88m/s$

Step 3:

The resultant speed in which the wreckage moves is

$v^2 = \sqrt{v_e^2 + v_n^2}$

$v^2 = \sqrt{(1.28)^2 + (-2.88)^2}$

$v^2 = \sqrt{1.6384 + 8.2944}$

$v^2 = \sqrt{9.9328}$

v = 3.15m/s

Step 4:

The direction of the wreckage is given as

tanθ = $\frac{v_n}{v_e}$

θ = $tan^{-1}(\frac{|-2.88|}{1.28} )$

θ = $tan^{-1}\frac{2.88}{1.28}$

θ = $tan^{-1} 2.25$

θ = 66.04°

Therefore, the wreckage begins to move with a speed of 3.15m/s in the direction 66.04°.

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