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Two billiard balls of equal mass undergo a perfectly elastic head-on collision. if the speed of ball 1 was initially 2.00 m/s, and that of ball 2 was 3.00 m/s in the opposite direction, what will be their speeds after the collision? ball 1

Respuesta :

skyluke89
The collision is elastic, this means that both total momentum and total kinetic energy are conserved after the collision.

- Let's start with conservation of momentum:
[tex]p_i = p_f[/tex]
Let's take as positive direction the initial direction of ball 1. Let's label with m the mass of the balls (they have same mass), with [tex]v_1[/tex] the final speed of ball 1, with [tex]v_2[/tex] the final speed of ball 2. The initial velocity of ball 1 is 2 m/s, while the initial velocity of ball 2 is -3 m/s (with a negative sign, because it is going in the opposite direction of ball 1). The conservation of momentum becomes
[tex](2 m/s) m - (3m/s)m = mv_1 + mv_2[/tex]
From which we find
[tex]v_2 = -1 m/s -v_1[/tex]

- Let's now write conservation of kinetic energy:
[tex]K_i = K_f[/tex]
which becomes
[tex] \frac{1}{2} m (2m/s)^2 + \frac{1}{2}m(3m/s)^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 [/tex]
Substituting 
[tex]v_2 = -1 m/s -v_1[/tex]
as we found in the conservation of momentum, and simplifying, we get
[tex]v_1^2 + v_1 - 6m/s = 0[/tex]

which has two solutions:
[tex]v_1 = -3 m/s, v_2 =2m/s[/tex] (1)
[tex]v_1=2 m/s, v_2 = -3 m/s[/tex] (2)
The second solution is meaningless: in fact, the balls have exactly same velocity and direction as before the collision. This would mean they didn't collide at all, so we can neglect solution 2. And so, the solution of the problem is (1).
onyebuchinnaji

The final velocity of the first ball is 3 m/s and the final velocity of the second ball is 2 m/s in opposite direction.

The given parameters;

  • mass of the billiard balls = m
  • initial speed of the first ball, u₁ = 2 m/s
  • initial speed of the ball u₂ = 3 m/s

Apply the principle of conservation linear momentum;

[tex]m_1u_1 \ + \ m_2u_2 = m_1v_1 + m_2v_2\\\\2m - 3m = mv_1 + mv_2\\\\-m = m(v_1 + v_2)\\\\-1 = v_1 + v_2[/tex]

Apply one-dimensional velocity;

[tex]u_1 +v_1 = u_2 + v_2\\\\2 + v_1 = -3 + v_2\\\\5 + v_1 = v_2[/tex]

From the first equation, the values of the final velocity is calculated as follows;

[tex]-1 = v_1 + 5 + v_1\\\\-1 = 2v_1 + 5\\\\-6 = 2v_1\\\\v_1 = -3 \ m/s[/tex]

then, v₂ = 5 - 3 = 2 m/s

Thus, the final velocity of the first ball is 3 m/s and the final velocity of the second ball is 2 m/s in opposite direction.

Learn more here:https://brainly.com/question/15869303

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