Ask question

Ask question

All categories

r-ruslan [8.4K]

2 years ago

3

Two carts are initially moving to the right on a low-friction track, with cart 1 behind cart 2. Cart 1 has a speed twice that of

cart 2 and so moves up and rear-ends cart 2, which has twice the inertia of cart 1. Part A Suppose that the the initial speed of cart 2 is v. What is the speed of cart 1 right after the collision if the collision is elastic?

1 answer:

Alexxx [7]2 years ago

4 0

Answer:

Explanation:

Given

cart 1 has twice the speed of cart 2

cart 2 has twice the mass of cart 1

Suppose m is the mass of cart 1 so mass of cart 2 is 2 m

suppose u is the initial velocity of car 2 so initial velocity of car 1 is u

For Elastic collision velocity after collision is given by

velocity of cart 2 is $v_2=\frac{2m_1}{m_1+m_2}\cdot u_1-\frac{m_1-m_2}{m_1+m_2}\cdot u_2$

velocity of cart 1 is $v_1=\frac{m_1-m_2}{m_1+m_2}\cdot u_1+\frac{2m_2}{m_1+m_2}\cdot u_2$

here $m_1=m,m_2=2 m$

$u_1=2u,u_2=u$

substituting values we get

$v_1=\frac{m-2m}{3m}\cdot 2u+2\cdot \frac{2m}{3m}\cdot u$

$v_1=\frac{2u}{3}$

$v_2=\frac{2m}{3m}\cdot (2u)-\frac{m-2m}{3m}\cdot u$

$v_2=\frac{5u}{3}$

You might be interested in

The sun transfers energy to the earth by radiation at a rate of approximately 1.00 kW per square meter of surface.

Mashutka [201]

Answer:

1320336992.2512 m²

1320.33 kilometers or 509.79 miles

Explanation:

Energy transferred by the sun

$W=0.24\times 1\times 10^3=240\ W/m^2$

Energy required by the United States is $1\times 10^{19}\ J/yr$ (assumed)

Power

$P=\frac{E}{t}\\\Rightarrow P=\frac{1\times 10^{19}}{365.25\times 24\times 3600}\\\Rightarrow P=316880878140.2895\ W$

Area

$A=\frac{P}{W}\\\Rightarrow A=\frac{316880878140.2895}{240}\\\Rightarrow A=132033699.2512\ m^2$

Area of the solar collector would be 1320336992.2512 m²

Converting to km²

$1\ m^2=\frac{1}{1000\times 1000}\ km^2$

$1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1000\times 1000}\ km^2=1320.33\ km^2$

Converting to mi²

$1\ m^2=\frac{1}{1609.34\times 1609.34}\ mi^2$

$1320336992.2512\ m^2=1320336992.2512\times \frac{1}{1609.34\times 1609.34}\ mi^2=509.79\ mi^2$

Each side of the square would be 1320.33 kilometers or 509.79 miles

4 0

2 years ago

Apply the impulse-momentum relation and the work-energy theorem to calculate the maximum value of t if the cake is not to end up

loris [4]

Puto chupame el semen ok? right?

8 0

2 years ago

A car travels three-quarters of the way around a circle of radius 20.0 m in a time of 3.0 s at a constant speed. the initial vel

schepotkina [342]

20.3 divided by 3.0 will get u velocity and v times 3.0s

3 0

2 years ago

The pail is rotated at a constant rate so it has the minimum speed at all points along its circular path. The water has mass m.

Alisiya [41]

Answer:

Explanation:

The pail is rotated at a constant rate in vertical circular path so it has the minimum speed at all points along its circular path . That means at top position the velocity is almost zero. In that case the centripetal force at top position will be provided by its weight or

mg = mv² / r ( r is radius of vertical circular path )

v = √ rg

At the bottom position its velocity will be increased due to loss of potential energy

so 1/2 m V² = 1/2 m v² + mg x 2r

V =√ 5 gr

If R be the reaction force at the bottom by bottom of pail

R - mg = mV² / r

R = mg +mV² / r

= mg + m x 5gr / r

R = 6mg

This is the magnitude of the force exerted by the water on the bottom of the pail .

7 0

2 years ago

A person is working on a steel structure while standing on the ground. An accident occurred where 5 A pass through the structure

matrenka [14]

Answer:

35mA

Explanation:

Hello!

To solve this problem we must use the following steps

1. Find the electrical resistance of the metal rod using the following equation

$R=\alpha \frac{l}{a}$

WHERE

α=

metal rod resistivity=2x10^-4 Ωm

l=leght=2m

A= Cross-sectional area

$A=\frac{\pi }{4} d^2=\frac{\pi }{4} (0.06)^2=0.00283$

solving

$R=(2x10^-4)\frac{2}{0.00283} =0.14$

2. Now we model the system as a circuit with parallel resistors, where we will call 1 the metal rod and 2 the man(see attached image)

3.we know that the sum of the currents in 1 and 2 must be equal to 5A, by the law of conservation of energy

I1+I2=5

4.as the voltage on both nodes is the same we can use ohm's law in resitance 1 and 2 (V=IR)

V1=V2

(0.14I1)=2000(i2)

solving for i1

I1=14285.7i2

5.Now we use the equation found in step 3

14285.7i2+i2=5

$i2=\frac{5}{14285.7+1} =3.5x10^-4A=35mA$

6 0

2 years ago