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2 years ago

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A truck of mass mT = 2000 kg is going north on Guadalupe with a speed of 4 m/s. The truck is struck by an eastbound car (mass of

1000 kg) traveling on 24th street. From the skidmarks after the collision, it is deduced that the mangled wreckage, which sticks together after the collision was skidding at angle 20◦ with respect to the initial direction of the car. What was the initial speed of the car?

1 answer:

pogonyaev2 years ago

5 0

Answer:21.97 m/s

Explanation:

Mass of truck $m_T=2000 kg$

Velocity of truck $v_c=4 m/s$

Mass of car $m_c=1000 kg$

let $v_c$ be the car velocity and u be the velocity of combined system after collision at angle of $20^{\circ}$

Conserving momentum in east and North direction Respectively

In east Direction

$m_c\cdot v_c=(m_c+m_T)u\cos 20$ -------1

In North direction

$m_T\cdot v_T=(m_c+m_T)u\sin 20$ ---------2

Divide 1 and 2 we get

$\frac{m_T\cdot v_T}{m_c\cdot v_c}=\frac{(m_c+m_T)u\sin 20}{(m_c+m_T)u\cos 20}$

$v_c=\frac{m_T\cdot v_T}{m_c\cdot \tan 20}$

$v_c=\frac{2000}{1000}\times \frac{4}{\tan 20}$

$v_c=21.97 m/s$

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$\Delta R=R_b-R_g\\\Rightarrow \Delta R=R_0(1+\alpha_b\Delta T)-R_0(1+\alpha_g\Delta T)\\\Rightarrow \Delta R=R_0(\alpha_b-\alpha_g)\Delta T\\\Rightarrow \Delta R=4\times (19\times 10^{-6}-9\times 10^{-6})\times (60-20)\\\Rightarrow \Delta R=0.0016\ cm$

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2 years ago

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Answer:

F = 69.3 N

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For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by

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