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2 years ago

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A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume

that any other unbalanced forces are negligible. (a) What is the speed of the wreckage just after the collision? (b) In what direction does the wreckage move just after the collision?

1 answer:

dalvyx [7]2 years ago

4 0

Explanation:

It is given that,

Mass of the car 1, $m_1=900\ kg$

Initial speed of car 1, $u_1=15i\ m/s$ (east)

Mass of the car 2, $m_2=750\ kg$

Initial speed of car 2, $u_1=20j\ m/s$ (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V$

$900\times 15i +750\times 20j=(900+750)V$

$13500i+15000j=1650V$

$V=(8.18i+9.09j)\ m/s$

The magnitude of speed,

$|V|=\sqrt{8.18^2+9.09^2}$

V = 12.22 m/s

(b) Let $\theta$ is the direction the wreckage move just after the collision. It is given by :

$tan\theta=\dfrac{v_y}{v_x}$

$tan\theta=\dfrac{9.09}{8.18}$

$\theta=48.01^{\circ}$

Hence, this is the required solution.

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