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3 years ago

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A 1022kg Caprice car stopped at an intersection is rear-ended by a 1620kg ranger truck moving with a speed of 14.5m/s. If the ca

r was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision.

1 answer:

Alika [10]3 years ago

8 0

Answer:

Explanation:

mass of car, m = 1022 kg

mass of truck, M = 1620 kg

initial velocity of truck, U = 14.5 m/s

initial velocity of car, u = 0 m/s

Let the final velocity of car is v and the final velocity of truck is V.

Collision is elastic, so the coefficient of restitution, e = 1

Use conservation of momentum

initial momentum of car + initial momentum of truck = final momentum of car + final momentum of truck

m x u + M x U = m x v + M x V

0 + 1620 x 14.5 = 1022 v + 1620 V

23490 = 1022 v + 1620 V ..... (1)

Use the formula of coefficient of restitution

$e = \frac{V_{1}-V_{2}}{u_{2}-u_{1}}$

1 (14.5 - 0) = v - V

14.5 = v - V

V = v - 14.5 .... (2)

Put in equation (1)

23490 = 1022 v + 1620 (v - 14.5)

23490 = 1022 v + 1620 v - 23490

46980 = 2642 v

v = 17.8 m/s

Put in equation (2)

V = 17.8 - 14.5

V = 3.3 m/s

Thus, the speed of car is 17.8 m/s and the velocity of truck is 3.3 m/s after collision.

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A object weighing 5 kg, starts to accelerate evenly on a horizontal line. A force moves the object

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3 years ago

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Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

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Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

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The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

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$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

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$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

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Answer:

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