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Fed [463]
3 years ago
8

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 of the spring. The mass of A is 2.00 times the mass of B, and the energy stored in the spring was 35 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:
Genrish500 [490]3 years ago
8 0

Explanation:

Given:

Final speed of mass A = Va

Final speed of mass B = Vb

Mass of A = Ma

Mass of B = Mb

Ma = 2 × Mb

By conservation of linear momentum,

0 = Ma × Va + Mb × Vb

0 =  2 × Mb × Va + Mb × Vb

Vb = - 2 × Va              

Energy of the spring, U = 1/2 × k × x^2

1/2 k x² = 1/2 × Ma × Va² + 1/2 × Mb × Vb²

35 = 1/2 × Ma × Va² + 1/2 × Mb × Vb²

Ma × Va² + Mb × Vb² = 70

2 × Mb (-Vb/2)² + Mb × Vb² = 70

1/2 × Mb × Vb² + Mb × Vb² = 70

3/2 × Mb × Vb² = 70

Mb × Vb² = 140/3

= 46.7 J

Ma = 2 × Mb and Vb = - 2 × Va

Ma/2 × (4 × Va²) = 140/3

Ma × Va² = 70/3

Kinetic energy of mass A, KEa = 1/2 × Ma × Va² = 23.3 J

Kinetic energy of mass B = 1/2 × Mb × Vb² = 46.7 J

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7.9x10^9 km is equal to

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Formula for electrostatic force is F = ( K q1 q2 )/ r²

where q1 and q2 represent the charges and r represents the distance between them and the value of K is 9 × 10⁹.

In question we have given

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value of q2 = 1.1 x 10⁻¹⁷ C

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Applying the formula

F = ( K x (5.4 x 10-7) x (1.1x10-17) )/ 0.25²

F = ( (9 × 10⁹ ) x (5.4 x 10-7) x (1.1x10-17) )/ 0.25²

F = ( (9 × 5.4 × 1.1) × ( 10⁹ × 10⁻⁷ x 10⁻¹⁷) )/ ( 6.25 × 10⁻² )

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So, The point charges are possessing equal and opposite electrostatic force of magnitude 8.5536 × 10⁻¹³ Newton.

Learn more about Electrostatic Force here:

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