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3 years ago

13

Applying the Law of Conservation of Momentum

1 answer:

IgorC [24]3 years ago

6 0

answer

the final velocity is half of train car B's initial velocity

explanation

the conservation of momentum states that the initial and final total momentum are equal

m1v1 + m2v2 for initial momentum

(m1 + m2)v3 for final momentum because both cars stick together so their masses are combined

m1v1 + m2v2 = (m1 + m2)v3

since the mass of both are the same, m1 = m2

m1v1 + m1v2 = (m1 + m1)v3

m1(v1 + v2) = 2m1v3

divide both sides by m1

v1 + v2 = 2v3

since train car A is initially at rest, v1 = 0

v2 = 2v3

v3 = v2 * 1/2

the final velocity is half of train car B's initial velocity

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Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m

Phoenix [80]

Answer:

a. $t_1=12.5\ s$

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c. $t_2=2.5714\ s$ is the time taken to stop after braking

Explanation:

Given:

speed of leading car, $u_1=25\ m.s^{-1}$

speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

deceleration of the leading car after braking, $a_1=-2\ m.s^{-2}$

a.

Time taken by the car to stop:

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where:

$v_1=0$ , final velocity after braking

$t_1=$ time taken

$0=25-2\times t_1$

$t_1=12.5\ s$

b.

using the eq. of motion for the given condition:

$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

$v_2=u_2+a_2.t_2$

$0=35-13.61\times t_2$

$t_2=2.5714\ s$ is the time taken to stop after braking

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Alexxandr [17]

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Mass of the soccer ball, m = 0.425 kg

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Part A,

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$\Delta p_x=mv\ cos\theta$

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$\Delta p_y=0.425\times 15\ sin(33)$

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Answer:

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Hence, the volume of air molecules is 1963.93 Moles

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3 years ago