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A 0.50-kg ball that is traveling at 6.0 m/s collides head-on with a 1.00-kg ball moving in the opposite direction at a speed of 12.0 m/s. The 0.50-kg ball bounces backward at 14 m/s after the collision. Find the speed of the second ball after the collision.

Respuesta :

skyluke89

Answer:

-2 m/s

Explanation:

Assuming the collision is elastic, the total momentum must be conserved:

[tex]p_i = p_f[/tex]

[tex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]

where

m1 = 0.50 kg is the mass of the first ball

m2 = 1.00 kg is the mass of the second ball

u1 = +6.0 m/s is the initial velocity of the first ball

u2 = -12.0 m/s is the initial velocity of the second ball

v1 = -14 m/s is the final velocity of the first ball

v2 = ? is the final velocity of the second ball

By re-arranging the equation, we can find the final velocity of the 1.00 kg ball:

[tex]v_2 = \frac{m_1 u_1 + m_2 u_2 - m_1 v_1}{m_2}=\frac{(0.50 kg)(6.0 m/s)+(1.00 kg)(-12 m/s)-(0.50 kg)(-14 m/s)}{1.00 kg}=-2 m/s[/tex]

which means, 2 m/s in the same direction the second ball was travelling before the collision.

asuwafohjames

The speed of the second ball is 2 m/s

To calculate the speed of the second ball after the collision, we use the formula below.

Formula:

  • mu+m'u' = mv+m'v'................. Equation 1

Where:

  • m = mass of the first ball
  • u = initial speed of the first ball
  • m' = mass of the second ball
  • u' = initial speed of the second ball
  • v = final speed of the first ball
  • v' = final speed of the second ball.

From the question,

Given:

  • m = 0.50 kg
  • m' = 1.00 kg
  • u = 6 m/s
  • u' = -12 m/s (opposite)
  • v = -14 m/s. (backward)

Substitute these values into equation 1

  • (0.5×6)+[1(-12)] = [0.5×(-14)]+(1×v')

Solve for v'

  • (3-12) = -7+v'
  • v'-7 = -9
  • v' = -9+7
  • v' = -2 m/s

Note: The negative sign can be ignored in the final answer

Hence, the speed of the second ball is 2 m/s.

Learn more about speed here: https://brainly.com/question/4931057

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