Physics, 20.01.2020 23:31 sierratellman

A7500 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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Answer from: kornut7316

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°.

A7500 kg truck traveling at 5m/s east collides with a 1500kg car moving at 20m/s in a direction 30 d

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Answer from: Quest

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