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timama [110]
3 years ago
12

Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t

he first having a mass of 140,000 kg and a velocity of 0.300 m/s, and the second having a mass of 95,000 kg and a velocity of −0.120 m/s. (The minus indicates direction of motion.) What is their final velocity (in m/s)?
Physics
2 answers:
bekas [8.4K]3 years ago
6 0

Answer:

0.13 m/s

Explanation:

m_1 = Mass of first car = 140000 kg

m_2 = Mass of second car = 95000 kg

u_1 = Initial Velocity of first car = 0.3 m/s

u_2 = Initial Velocity of second car = -0.12 m/s

v = Velocity of combined mass

For elastic collision

m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow v=\frac{m_1u_1 + m_2u_2}{m_1 + m_2}\\\Rightarrow  v=\frac{140000\times 0.3 + 95000\times -0.12}{140000+ 95000}\\\Rightarrow v=0.13\ m/s

Their final velocity is 0.13 m/s

Licemer1 [7]3 years ago
4 0

Answer:

The final velocity of the two train cars is 0.13 m/s.

Explanation:

Given that,

Mass of the first car, m_1=14000\ kg

Mass of the second car, m_2=95000\ kg

Initial speed of first car, u_1=0.3\ m/s

Initial speed of second car, u_2=-0.12\ m/s

It is mentioned that two train cars are coupled together by being bumped into one another. So, it is a case of inelastic collision. Momentum will remain conserved here. Using the conservation of linear momentum we get :

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

V is the final speed of two cars.

V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{140000\times 0.3+95000\times (-0.12)}{(140000+95000)}\\\\V=0.13\ m/s

So, the final velocity of the two train cars is 0.13 m/s. Hence, this is the required solution.

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Starting from rest, a runner reaches a speed of 2.8 m/s in 2.1 s. In the same time 2.1 s time, a motorcycles increases speed fro
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a) Acceleration of runner is 1.33 m/s²

b)  Acceleration of motorcycle is 2.85 m/s²

c) The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.

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t = Time taken

u = Initial velocity = 0

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a = Acceleration

Equation of motion

v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{2.8-0}{2.1}\\\Rightarrow a=1.33\ m/s^2

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v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{43-37}{2.1}\\\Rightarrow a=2.85\ m/s^2

Acceleration of motorcycle is 2.85 m/s²

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{2.8^2-0^2}{2\times 1.33}\\\Rightarrow s=2.94\ m

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v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{43^2-37^2}{2\times 2.85}\\\Rightarrow s=84.21\ m

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A typical laboratory centrifuge rotates at 4000 rpm. Test tubes have to be placed into a centrifuge very carefully because of th
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Answer:

1. a_{rad}=17545.2\frac{m}{s^{2}}

2. a=4429.45 \frac{m}{s^{2}}

Explanation:

Radial acceleration is:

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With r the radius respects the axis of rotation and v the tangential velocity that is related with angular velocity (ω) by:

v= \omega r (2)

By (2) on (1):

a_{rad}= \frac{(\omega r)^2}{r}= (\omega )^2r=(418,88\frac{rad}{s})^2(0.1m)

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To find the acceleration of the tube with the fall, we can use the expression:

\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t) (3)

Due impulse-momentum theorem:

\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i} (4)

with p the momentum and J the impulse. By (4) on (3):

\overrightarrow{p}_{f}-\overrightarrow{p}_{i}=\overrightarrow{F}_{avg}(\varDelta t)

And using Newton's second law (F=ma) and that (P=mv):

mv_f-mv_i=(ma)(\varDelta t) (5)

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\frac{mv_i}{2}=mgh

v_i=\sqrt{2gh}=\sqrt{2(9.81)(1.0)}=4.43\frac{m}{s}

So (5) is:

-m(4.43)=(ma)(\varDelta t)

solving for a:

a=\frac{4.43}{\varDelta t}=\frac{4.43}{1.0\times10^{-3}}=-4429.45

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