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Artist 52 [7]
3 years ago
11

a 2kg marble is moving at 3 m/s when it strikes another 4kg marble moving in the opposite direction at -3 m/s. What will be the

velocity of the 4kg marble after the collision of the 2kg marble's new velocity is -4 m/s. Also state the direction of the 2kg marble after they hit.
Physics
1 answer:
Elan Coil [88]3 years ago
4 0
Momentum is conserved after the collision 

Momentum of 2 Kg before collision = 2 * 3 = 6
Momentum of 4 kg  before collision = 4 * -3 = -12

so 6 + -12 = 2 * -4 + 4 *x   where x is velocity of 4kg marble.

4x - 8 = -6
4x = 2
x = 0.5 

Velocity of 4 kg marble is 0.5 m/s after collision

The 2 kg marble will move in the opposite direction to which it was moving before the collision.
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Answer:

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Equivalences

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known data

q₁=-2.9nC=-2.9 *10⁻⁹C

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Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2}  } =\frac{8.99*10^{9}*2.9*10^{-9}  }{0.84^{2} } =36.95\frac{N}{C}

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Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2}  } =-21.4\frac{N}{C}

Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2}  } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2}  } =-17.11\frac{N}{C}

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

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