Physics, 28.12.2019 22:31 Marqiuse412

A900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s. the cars stick together after the collision. in what direction does the wreckage move just after the collision?

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Answer from: nolam123

The wreckage moves in the direction 48° N of E just after the collision

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Further explanation

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

$\large {\boxed {F = ma }$

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

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Given:

mass of car 1 = m₁ = 900 kg

initial velocity of car 1 = u₁ = 15 i m/s

mass of car 2 = m₂ = 750 kg

initial velocity of car 2 = u₂ = 20 j m/s

Asked:

final velocity = v = ?

Solution:

Firstly , we will use Conservation of Momentum Law as follows:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

$900(15 \widehat{i}) + 750(20 \widehat{j}) = 900v + 750v$

$13500 \widehat{i} + 15000 \widehat{j} = 1650v$

$v = (13500 \widehat{i} + 15000 \widehat{j}) \div 1650$

$v = \frac{90}{11} \widehat{i} + \frac{100}{11} \widehat{j}$

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Next , we could find the direction of this final velocity as follows:

$\theta = \tan^{-1} \frac{v_y}{v_x}$

$\theta = \tan^{-1} \frac{100/11}{90/11}$

$\theta = \tan^{-1} \frac{10}{9}$

$\theta \approx 48^o \texttt{ N of E}$

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Learn moreImpacts of Gravity : Effect of Earth’s Gravity on Objects : The Acceleration Due To Gravity : Newton's Law of Motion: Example of Newton's Law:

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

A900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m

Answer from: Quest

Inelastic is the answer your looking for.

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge