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Gnesinka [82]
4 years ago
14

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

m apart, and they then fly off in opposite directions, free from the spring. The mass of A is 2.00 times the mass of B, and the energy stored in the spring was 60 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of (a) particle A and (b) particle B?
Physics
1 answer:
Alinara [238K]4 years ago
4 0

Answer:

Kinetic energy of mass A = 20 J

Kinetic energy of mass B = 40 J

Explanation:

Lets take final speed of mass A  = v

The final speed of mass B = v'

mass of A= m

mass of B = m'

m = 2 m'

There is no any external force so the linear momentum will be conserve.

Pi = pf

0 = m v + m ' v'

0 =  2 m ' v +m ' v'

v ' = - 2 v                          

Now from energy conservation

1/2 k x²=1/2 m v² + 1/2 m' v'²

60 = 1/2 m v² + 1/2 m' v'²

m v² +  m' v'² = 120

2 m ' (-v'/2)² + m' v'² = 120

m ' v'²/2 + m' v'² = 120

m' v '² = 80

m =2 m'  and v ' = - 2 v

So

m/2 ( 4 v²) = 80

m v ²=40

So the kinetic energy of mass A= 1/2 m v² = 20 J

Kinetic energy of mass B = 1/2 m' v'² = 40 J

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Two positive charges of magnitude 20 micro C and 100 micro C are placed at a distance of 150cm. Calculate the force of repulsion
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first \: positive \: charge = q1 = 20

1micro \: =   {1}^{ - 6}

1q = 20 \times 10 {}^{ - 6}

second \: charge = q2 = 100 \times 10 {}^{ - 6}

formula = f =  \frac{k \times q1 × q2}{d {}^{2} }

remember

k = 9 \times  {10}^{9}

change distance 150cm to 1.5m

putting values

f =\frac{9 \times 10 {}^{9} \times 20  \times 10 {}^{ - 6} \times 100 \times  {10}^{ - 6}   }{1.5 {}^{2} }

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A stone falls freely from rest for 8.0s what is it final velocity
NemiM [27]
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7 0
3 years ago
Two small conducting point charges, separated by 0.4 m, carry a total charge of 200 C. They repel one another with a force of 12
Lunna [17]

Answer:

200 C

Explanation:

Let C1 and C2 be their charges. According to Coulomb's law

F_C = k\frac{C_1C_2}{R^2}

where k = 8.99\times10^9 nm^2/C^2 is the constant, R = 0.4m is the distance between them, F = 120 N is their resulting charge force

120 = 8.99\times10^9\frac{C_1C_2}{0.4^2}

C_1C_2 = \frac{120*0.4^2}{8.99\times10^9} = 2.13\times10^{-9}

Since their total charge is 200C:

C_1 + C_2 = 200 or C_1 = 200 - C_2

We can substitute the above equation

C_1C_2 = (200 - C_2)C_2 = 2.13\times10^{-9}

-C_2^2 +200C_2 - 2.13\times10^{-9} = 0

C= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

C= \frac{-200\pm \sqrt{(200)^2 - 4*(-1)*(-0.00000000213)}}{2*(-1)}

C= \frac{-200\pm200}{-2}

C = 1.06 \times 10^{-11} or C \approx 200

So the larger charge is C = 200 C

8 0
4 years ago
How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to
finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is x=\sqrt{\frac{2\times EPE}{k}}

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

EPE=\frac{1}{2}kx^2

Rewriting the above expression in terms of 'x', we get:

x=\sqrt{\frac{2\times EPE}{k}}

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0
3 years ago
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