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nika2105 [10]
4 years ago
8

A car moving with an initial speed v collides with a second stationary car that is one-half as massive. After the collision the

first car moves in the same direction as before with a speed v/3.(a) Find the final speed of the second car.(b) Is this collision elastic or inelastic?
Physics
1 answer:
Arisa [49]4 years ago
3 0

(a) \frac{4}{3}v

Let's write the law of conservation of momentum for the collision:

m_A u_A + m_B u_B = m_A v_A + m_B v_B (1)

where u refers to the initial velocities and v refers to the velocity after the collision.

The problem gives us the following information:

- car B is one-half as massive as car A, so we can write:

m_A = 2m\\m_B = m

- car B is initiall stationary:

u_B=0

- After the collision, car A moves in the same direction as before with a speed v/3:

u_A = v\\v_A = \frac{1}{3}u_A=\frac{1}{3}v

So we can rewrite (1) as

(2m) v = (2m) \frac{v}{3}+mv_B

and solving for v_B, we find the final speed of car B:

v_B = \frac{6v-2v}{3}=\frac{4}{3}v

(b) Elastic

To find if the collision is elastic or inelastic, we have to check if the total kinetic energy has been conserved or not.

The total kinetic energy before the collision is:

K_i = \frac{1}{2}m_A u_A^2 = \frac{1}{2}(2m)(v)^2=mv^2

The total kinetic energy after the collision is:

K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2=\frac{1}{2}(2m)(\frac{v}{3})^2+\frac{1}{2}m(\frac{4}{3}v)^2=\frac{1}{9}mv^2+\frac{8}{9}mv^2=mv^2

The total kinetic energy has been conserved: so, the collision is elastic.

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<h2>Answer:</h2>

<h2>Explanation:</h2>

First, let's refer to the distance formula:

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<h3>1. Speed 1.</h3>

In order to use the formula, we need to convert minutes into hours since the speed is given in km/h.

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<h3>2. Speed 2.</h3>

Convert minutes to hours again and do the same calculations.

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<h3>3. Speed 3.</h3>

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<h3>4. Obtain the total distance.</h3>

The total distance must be given by the addition of all individual distances traveled by the car on each speed:

Total distance= 30.38 km + 19.08 km + 18.85 km= 68.31 km.

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